2Trigonometry UNCORRECTED PAGE PROOFS. 2.1 Kick off with CAS

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1 . Kick off with CAS Trigonometr. Reciprocal trigonometric functions. Trigonometric identities using reciprocal trigonometric functions. Compound-angle formulas.5 Double-angle formulas. Inverse trigonometric functions.7 General solutions of trigonometric equations.8 Graphs of reciprocal trigonometric functions.9 Graphs of inverse trigonometric functions.0 Review

2 . Kick off with CAS Eploring inverse trigonometric functions In this topic, we will investigate the inverse trigonometric functions. Using CAS, determine each of the following. Remember to have the calculator in radians mode. a cos cos 5 b cos (cos()) c cos (cos()) d cos cos 7 e cos cos 8 7 f cos cos g cos (cos( ) ) For what values of is cos (cos() ) =? Confirm our result using CAS. Using CAS, determine each of the following. a tan tan b tan tan 5 c tan (tan() ) d tan tan 5 e tan tan 7 5 f tan tan g tan tan For what values of is tan (tan()) =? Confirm our result using CAS. Please refer to the Resources tab in the Prelims section of our ebookplus for a comprehensive step-b-step guide on how to use our CAS technolog.

3 . Units & AOS Topic Concept Reciprocal circular functions Concept summar Practice questions Reciprocal trigonometric functions Histor of trigonometr The word trigonometr is derived from the Greek words trigonon and metron, meaning triangle and measure. Trigonometr is the branch of mathematics that deals with triangles and the relationships between the angles and sides of a triangle. Trigonometr was originall devised in the third centur BC to meet the needs of the astronomers of those times. Hipparchus was a Greek astronomer and mathematician and is considered to be the founder of trigonometr, as he compiled the first trigonometric tables in about 50 BC. Definitions of trigonometric ratios The following is a review of trigonometr, which is needed for the rest of this topic and subsequent work in this book. The trigonometric functions sin(), cos() and tan() are defined in terms of the ratio of the lengths of the sides of a right-angled triangle. Let the lengths of the three sides of the triangle be a, b and c, and let the angle between sides a and c be θ. sin(θ) = opposite hpotenuse = b c cos(θ) = adjacent hpotenuse = a c tan(θ) = opposite adjacent = b a Pthagoras theorem states that in an right-angled triangle, the square of the hpotenuse is equal to the sum of the squares of the other two sides. That is: a + b = c. The unit circle An alternative definition of the trigonometric functions is based on the unit circle, which is a circle with radius one unit and centre at the origin. The unit circle has the equation + =. The coordinate of an point P (, ) on the unit circle is defined in terms of the trigonometric functions OR = = cos(θ) and RP = = sin(θ), where θ is the angle measured as a positive angle, anticlockwise from positive direction of the -ais. The trigonometric functions are also called circular functions as the are based on the unit circle. θ (0, ) θ = T P tan (θ) (, 0) θ (, 0) θ = O R S θ = 0 (0, ) c a θ = b 58 Maths Quest SPECIALIST MATHEMATICS VCE Units and

4 B substituting = cos(θ) and = sin(θ) into the equation + =, we can derive the relationship sin (θ) + cos (θ) =. Note that sin (θ) = (sin(θ)) and cos (θ) = (cos(θ)). The vertical distance from S to T is defined as tan(θ). As the triangles ΔORP and ΔOST are similar, RP OR = ST OS = tan(θ) = sin(θ) cos(θ) = tan(θ) Angles of an magnitude In the diagram of the unit circle, consider the point (0, ) on the -ais. This point corresponds to the angle θ = 90 or radians rotated from the positive end of the -ais. Since the sine of the angle is the -coordinate, it follows that sin(90 ) = sin =. Since the cosine of the angle is the -coordinate, it follows that cos (90 ) = cos = 0. The tangent is the value of sine divided b the cosine; because we cannot divide b zero, the tan of θ = 90 or radians is undefined. Similarl for the point (, 0), where θ = 80 or radians, it follows that cos(80 ) = cos() = and sin(80 ) = sin() = 0. The diagram can be used to obtain the trigonometric value of an multiple of 90, and these results are summarised in the following table. Angle (degrees) Angle (radians) 0 sin (θ) cos (θ) 0 0 tan (θ) 0 Undefined 0 Undefined 0 Note: Whenever an angle measurement is shown without a degree smbol in this topic, assume that it is measured in radians. The first quadrant The angle in the first quadrant is 0 < θ < 90 in degrees or 0 < θ < in radians. In the first quadrant, > 0 and > 0, so cos(θ) > 0 and sin(θ) > 0; therefore, tan(θ) > 0. The following table shows values derived from triangles in the first quadrant using the trigonometric ratios. You should memorise these values, as the are used etensivel in this topic. Topic Trigonometr 59

5 Angle (degrees) Angle (radians) 0 sin (θ) 0 cos (θ) 0 tan (θ) 0 Note that sin(0 ) + sin(0 ) sin(90 ) and in general sin(a + B) sin(a) + sin(b), cos(a + B) cos(a) + cos(b) and tan(a + B) tan(a) + tan(b). Undefined The formulas for sin(a + B) are called compound-angle formulas. The are studied in greater depth in Section.. The second quadrant The angle in the second quadrant is 90 < θ < 80 in degrees or P (a, b) < θ < in radians. In the second Pʹ ( a, b) θ quadrant, < 0 and > 0, so cos(θ) < 0 and b b sin(θ) > 0; therefore, tan(θ) < 0. θ a Consider the point P (a, b) in the first quadrant. O a When this point is reflected in the -ais, it becomes the point P ( a, b). If P makes an angle of θ with the -ais, then P makes an angle of 80 θ degrees or θ radians with the -ais. From the definitions of sine and cosine, we obtain the following relationships. sin(80 θ) = sin(θ) sin( θ) = sin(θ) cos(80 θ) = cos(θ) cos( θ) = cos(θ) tan(80 θ) = tan(θ) tan( θ) = tan(θ) For eample: sin 5 = sin = sin = cos = cos = cos = tan = tan = tan = The third quadrant The angle in the third quadrant is 80 < θ < 70 in degrees or < θ < in radians. In the third quadrant, < 0 and < 0, so cos(θ) < 0 and sin(θ) < 0. However, tan(θ) > 0. T Tʹ tan ( θ) 0 Maths Quest SPECIALIST MATHEMATICS VCE Units and

6 Consider the point P (a, b) in the first quadrant. When this point is reflected in both the - and -aes, it becomes the point P ( a, b). If P makes an angle of θ with the -ais, then P makes an angle of 80 + θ degrees or + θ radians with the positive end of the -ais. From the definitions of sine and cosine, we obtain the following relationships. sin(80 + θ) = sin(θ) sin( + θ) = sin(θ) cos(80 + θ) = cos(θ) cos( + θ) = cos(θ) tan(80 + θ) = tan(θ) tan( + θ) = tan(θ) For eample: sin cos tan 7 5 = sin + = sin = = cos + = cos = = tan + = tan = The fourth quadrant The angle in the fourth quadrant is 70 < θ < 0 in degrees or P (a, b) < θ < in radians. In the fourth quadrant, > 0 and < 0, so b cos(θ) > 0 and sin(θ) < 0; therefore, tan(θ) < 0. θ Consider the point P (a, b) in the first quadrant. O a θ b When this point is reflected in the -ais, it becomes the point P (a, b). If P makes an angle of θ with the -ais, then P makes an angle of Pʹ (a, b) 0 θ degrees or θ radians with the -ais. From the definitions of sine and cosine, we obtain the following relationships. sin(0 θ) = sin(θ) sin( θ) = sin(θ) cos(0 θ) = cos(θ) cos( θ) = cos(θ) tan(0 θ) = tan(θ) tan( θ) = tan(θ) For eample: 7 sin = sin = sin = cos tan 5 b Pʹ ( a, b) = cos = cos = + θ a = tan = tan = P (a, b) θ O a b T tan ( + θ) T Tʹ tan (θ) Topic Trigonometr

7 Summar The trigonometric ratios sin(θ), cos(θ) and tan(θ) are all positive in the first quadrant. Onl sin(θ) is positive in the second quadrant; onl tan(θ) is positive in the third quadrant; and finall, onl cos(θ) is positive in the fourth quadrant. This is summarised in the diagram at right. The mnemonic CAST is often used as a memor aid. sin(θ) = sin( θ) = sin( + θ) = sin( θ) cos(θ) = cos( θ) = cos( + θ) = cos( θ) tan(θ) = tan( θ) = tan( + θ) = tan( θ) Negative angles A negative angle is one that is measured clockwise from the positive direction of the -ais. Consider the point P (a, b) in the first quadrant. When this point is reflected in the -ais, it becomes the point P (a, b). If P makes an angle of θ with the -ais, then P makes an angle of θ with the -ais. From the definitions of sine and cosine, we obtain the following relationships. sin( θ) = sin(θ) cos( θ) = cos(θ) tan( θ) = tan(θ) A negative angle < θ < 0 is just the equivalent angle in the fourth quadrant. For positive angles greater than 0 or, we can just subtract multiples of 0 or. sin(0 + θ) = sin(θ) sin( + θ) = sin(θ) cos(0 + θ) = cos(θ) cos( + θ) = cos(θ) tan(0 + θ) = tan(θ) tan( + θ) = tan(θ) For eample: sin cos 7 tan = sin = cos = tan 7 = sin + = sin = = cos = cos = = tan = tan = Reciprocal trigonometric functions The reciprocal of the sine function is called the cosecant function, often abbreviated to cosec. It is defined as cosec() =, provided that sin() 0. sin() θ θ = + θ S T O θ = A C θ = P (a, b) b θ θ a b Pʹ (a, b) θ θ = 0 θ T tan (θ) Tʹ Maths Quest SPECIALIST MATHEMATICS VCE Units and

8 WorKeD example The reciprocal of the cosine function is called the secant function, often abbreviated to sec. It is defined as sec() =, provided that cos() 0. cos() The reciprocal of the tangent function is called the cotangent function, often abbreviated to cot. It is defined as cot() = tan() = cos(), provided that sin() 0. sin() Note that these are not the inverse trigonometric functions. (The inverse trigonometric functions are covered in Section..) The reciprocal trigonometric functions can also be defined in terms of the sides of a right-angled triangle. cosec (θ) = hpotenuse = c opposite b sec (θ) = hpotenuse = c c b adjacent a cot (θ) = adjacent opposite = a b θ a eact values The eact values for the reciprocal trigonometric functions for angles that are multiples of 0 and 5 can be found from the corresponding trigonometric values b finding the reciprocals. Often it is necessar to simplif the resulting epression or rationalise the denominator. Find the eact value of cosec 5. WritE State the required identit. cosec (θ) = sin (θ) 5 cosec = 5 sin Use the known results. Use sin ( + θ) = sin (θ) with θ =. Simplif the ratio and state the final answer. cosec cosec 5 5 = sin + = sin = = Topic TrIgonoMeTr

9 WorKeD example using triangles to fi nd values Triangles can be used to find the values of the required trigonometric ratios. Particular attention should be paid to the sign of the ratio. If cosec (θ) = 7 and < θ <, find the eact value of cot(θ). State the values of the sides of a corresponding right-angled triangle. WritE/draW cosec (θ) = 7 sin(θ) = 7 sin(θ) = 7 The hpotenuse has a length of 7 and the opposite side length is. Draw the triangle and label the side lengths using the definition of the trigonometric ratio. Label the 7 unknown side length as. Calculate the value of the third side using Pthagoras theorem. State the value of a related trigonometric ratio. 5 Calculate the value of the required trigonometric value. + = 7 + = 9 = 9 = = Given that < θ <, θ is in the second quadrant. Although sin(θ) is positive in this quadrant, tan(θ) is negative. tan(θ) = cot (θ) = tan (θ) = = 7 θ θ MaTHs QuesT specialist MaTHeMaTICs VCe units and

10 Eercise. PRactise Consolidate Reciprocal trigonometric functions WE Find the eact value of cosec Find the eact value of sec 7.. WE If cosec (θ) = 5 and < θ <, find the eact value of cot (θ). If cot (θ) = and < θ <, find the eact value of sec (θ). 5 Find the eact values of each of the following. a sec b sec Find the eact values of each of the following. a cosec b cosec 5 7 Find the eact values of each of the following. a cot b cot c sec c cosec c cot a If sin() = and < <, find the eact value of sec(). b If cosec() = and < <, find the eact value of cot(). 9 a If cos() = 7 and < <, find the eact value of cot(). b If sec() = 5 and < <, find the eact value of cosec(). 0 a If cos() = 7 and b If sec() = 8 5 and < <, find the eact value of cosec(). < <, find the eact value of cot(). a If cosec() = and < <, find the eact value of tan(). b If cot() = 5 and < <, find the eact value of cosec(). a If sec() = 7 and < <, find the eact value of cot(). b If cot() = and < <, find the eact value of cosec(). a If sec() = and b If cot() = 5 and < <, find the eact value of cosec(). < <, find the eact value of sec (). d sec 7 d cosec d cot 5 Topic Trigonometr 5

11 MastEr. Units & AOS Topic Concept Trigonometric identities Concept summar Practice questions WorKeD example a If cot() = and < <, find the eact value of cosec(). b If sec() = and < <, find the eact value of cot(). 5 If cosec() = p q where p, q R+ and < <, evaluate sec() cot(). If sec() = a b where a, b R+ and < <, evaluate cot() cosec(). Trigonometric identities using reciprocal trigonometric functions Identities B mathematical convention, (sin(θ)) is written as sin (θ), and similarl (cos(θ)) is written as cos (θ). Note that sin (θ) + cos (θ) = is an identit, not an equation, since it holds true for all values of θ. Similarl, tan(θ) = sin(θ) holds for all values of θ for which tan(θ) is defined, that is cos(θ) for all values where cos(θ) 0, or θ (n + ) where n Z or odd multiples of. Proving trigonometric identities A trigonometric identit is verified b transforming one side into the other. Success in verifing trigonometric identities relies upon familiarit with known trigonometric identities and using algebraic processes such as simplifing, factorising, cancelling common factors, adding fractions and forming common denominators. The following identities must be known. tan(θ) = sin(θ) cos(θ) sec(θ) = cos(θ) cosec(θ) = sin(θ) cot(θ) = tan(θ) Prove the identit tan (θ) + cot (θ) = sec (θ)cosec (θ). WritE Start with the left-hand side. LHS = tan(θ) + cot(θ) Substitute for the appropriate trigonometric identities. tan(θ) = sin(θ) cos(θ) LHS = sin(θ) cos(θ) + cos(θ) sin(θ) and cot(θ) = cos(θ) sin(θ) MaTHs QuesT specialist MaTHeMaTICs VCe units and

12 Add the fractions, forming the lowest common denominator. LHS = sin (θ) + cos (θ) cos(θ)sin(θ) Simplif the numerator. Since sin (θ) + cos (θ) =, LHS = cos(θ)sin(θ) 5 Write the epression as factors. LHS = cos(θ) sin(θ) Substitute for the appropriate trigonometric identities. The proof is complete. WorKeD example Fundamental relations If all terms of sin (θ) + cos (θ) = are divided b sin (θ), we obtain sin (θ) sin (θ) + cos (θ) sin (θ) = and hence obtain the trigonometric identit sin (θ) + cot (θ) = cosec (θ). If all terms of sin (θ) + cos (θ) = are divided b cos (θ), we obtain sin (θ) cos (θ) + cos (θ) cos (θ) = and hence obtain the trigonometric identit cos (θ) tan (θ) + = sec (θ). Prove the identit + tan (θ) + cot (θ) = tan (θ). WritE Start with the left-hand side. LHS = + tan (θ) + cot (θ) Substitute the appropriate trigonometric identities. Use appropriate trigonometric identities to epress the quotient in terms of sines and cosines. sec(θ) = and cosec(θ) = cos(θ) sin(θ) LHS = sec(θ)cosec(θ) = RHS Replace + tan (θ) = sec (θ) in the numerator and + cot (θ) = cosec (θ) in the denominator. LHS = sec (θ) cosec (θ) sec (θ) = cos (θ) and cosec (θ) = sin (θ) cos (θ) LHS = sin (θ) Topic TrIgonoMeTr 7

13 Simplif the quotient. Use a = b a. b LHS = sin (θ) cos (θ) 5 Simplif and state the final result. Since tan (θ) = sin (θ) cos (θ), Eercise. PRactise Consolidate LHS = tan (θ) = RHS Trigonometric identities using reciprocal trigonometric functions WE Prove the identit sec (θ) + cosec (θ) = sec (θ)cosec (θ). sin(θ) Prove the identit + cos(θ) + + cos(θ) = cosec(θ). sin(θ) WE Prove the identit + cot (θ) + tan (θ) = cot (θ). Prove the identit ( sin (θ))( + tan (θ)) =. For questions 5, prove each of the given identities. 5 a cos(θ)cosec(θ) = cot(θ) b cos(θ)tan(θ) = sin(θ) a sin(θ)sec(θ)cot(θ) = b cos(θ)cosec(θ)tan(θ) = 7 a (cos(θ) + sin(θ)) + (cos(θ) sin(θ)) = b cos (θ) = sin (θ) 8 a tan (θ)cos (θ) + cot (θ)sin (θ) = sin(θ) b cosec(θ) + cos(θ) sec(θ) = 9 a sin(θ) + + sin(θ) = sec (θ) b cos(θ) + + cos(θ) = cosec (θ) 0 a + sec (θ) + + cos (θ) = b ( tan(θ)) + ( + tan(θ)) = sec (θ) a (tan(θ) + sec(θ)) = + sin(θ) sin(θ) b sec (θ) sec (θ) = tan (θ) + tan (θ) tan(θ) a sec(θ) + tan(θ) sec(θ) + = cosec(θ) b + cot(θ) sec(θ) cosec(θ) tan(θ) + cot(θ) = cos(θ) 8 Maths Quest SPECIALIST MATHEMATICS VCE Units and

14 Master. Units & AOS Topic Concept Compound- and double-angle formulas Concept summar Practice questions a a cos(θ) sin(θ) = sec(θ) + tan(θ) b cos(θ) = sec(θ) tan(θ) + sin(θ) + sin (θ) + + cosec (θ) = b + cot (θ) + + tan (θ) = For questions 5 and, prove each of the given identities. 5 a a b cos (θ) = b + (a b)cosec (θ) b a b sin (θ) = b + (a b)sec (θ) sin (θ) cos (θ) a a b tan (θ) = (a + b)cos (θ) b b a b cot (θ) = (a + b)sin (θ) b + tan (θ) + cot (θ) Compound-angle formulas The compound-angle formulas are also known as trigonometric addition and subtraction formulas. Proof of the compound-angle formulas The compound addition formulas state that: sin(a + B) = sin(a)cos(b) + cos(a)sin(b) cos(a + B) = cos(a)cos(b) sin(a)sin(b) tan(a) + tan(b) tan(a + B) = tan(a)tan(b) It is interesting to consider one method of proving these identities. Consider the triangle OQR with a right angle at Q, as shown in the diagram. The line segment TR is constructed so that TR is perpendicular to OR, and the line segment TP is constructed so that it is perpendicular to OP and SR. Let ROQ = A and TOR = B so that TOP = A + B. O Using the properties of similar triangles in ΔTSR and ΔOQR, or the propert that supplementar angles sum to 90, it follows that STR = A. In triangle OQR, sin(a) = QR OQ and cos(a) = OR OR. In triangle RST, sin(a) = SR ST and cos(a) = RT RT. In triangle ORT, sin(b) = RT OR and cos(b) = OT OT. Now consider the triangle OPT. sin(a + B) = PT PS + ST = = PS OT OT OT + ST OT B A S T P Q R Topic Trigonometr 9

15 PS = QR, so That is, Also in the triangle OPT: PQ = SR, so sin(a + B) = QR OT + ST OT = QR OT OR OR + ST OT RT RT = QR OR OR OT + ST RT RT OT sin(a + B) = sin(a)cos(b) + cos(a)sin(b) cos(a + B) = OP OQ PQ = = OQ OT OT OT PQ OT cos(a + B) = OQ OT SR OT = OQ OT OR OR SR OT RT RT = OQ OR OR OT SR RT RT OT cos(a + B) = cos(a)cos(b) sin(a)sin(b) Proof of the compound-angle subtraction formulas The compound subtraction formulas state that: sin(a B) = sin(a)cos(b) cos(a)sin(b) cos(a B) = cos(a)cos(b) + sin(a)sin(b) These formulas can obtained b replacing B with B and using cos( B) = cos(b) and sin( B) = sin(b). Substituting into the formula sin(a + B) = sin(a)cos(b) + cos(a)sin(b), we derive sin(a + ( B)) = sin(a)cos( B) + cos(a)sin( B), so that sin(a B) = sin(a)cos(b) cos(a)sin(b). Similarl, in the formula cos(a + B) = cos(a)cos(b) sin(a)sin(b), we derive cos(a + ( B)) = cos(a)cos( B) sin(a)sin( B), so that cos(a B) = cos(a)cos(b) + sin(a)sin(b). Proof of the compound-angle formulas involving tangents Let us substitute the formulas for sin(a + B) and cos(a + B) into the identit for the tangent ratio. sin(a + B) tan(a + B) = cos(a + B) sin(a)cos(b) + cos(a)sin(b) = cos(a)cos(b) sin(a)sin(b) 70 Maths Quest SPECIALIST MATHEMATICS VCE Units and

16 WorKeD example 5 In order to simplif this fraction, divide each term b cos(a)cos(b): sin(a)cos(b) cos(a)cos(b) + cos(a)sin(b) cos(a)cos(b) tan(a + B) = cos(a)cos(b) cos(a)cos(b) sin(a)sin(b) cos(a)cos(b) = sin(a) cos(a) + sin(b) cos(b) sin(a) cos(a) sin(b) cos(b) tan(a) + tan(b) = tan(a)tan(b) The corresponding formula for the tangent for the difference of two angles is obtained b replacing B with B and using tan( B) = tan(b). tan(a) + tan( B) tan(a + ( B)) = tan(a)tan( B) tan(a B) tan(a) tan(b) = + tan(a)tan(b) summar of the compound-angle formulas These results are called the compound-angle formulas or addition theorems. The can be summarised as: sin(a + B) = sin(a)cos(b) + cos(a)sin(b) sin(a B) = sin(a)cos(b) cos(a)sin(b) cos(a + B) = cos(a)cos(b) sin(a)sin(b) cos(a B) = cos(a)cos(b) + sin(a)sin(b) tan(a) + tan(b) tan(a + B) = tan(a)tan(b) tan(a) tan(b) tan(a B) = + tan(a)tan(b) using compound-angle formulas in problems The compound-angle formulas can be used to simplif man trigonometric epressions. The can be used in both directions, for eample sin(a)cos(b) + cos(a)sin(b) = sin(a + B). Evaluate sin ( )cos (8 ) + cos ( )sin (8 ). WritE State an appropriate identit. sin(a)cos(b) + cos(a)sin(b) = sin(a + B) Let A = and B = 8. sin( )cos(8 ) + cos( )sin(8 ) = sin( + 8 ) Topic TrIgonoMeTr 7

17 Simplif and use the eact values. sin ( )cos (8 ) + cos ( )sin (8 ) = sin (0 ) WorKeD example epanding trigonometric epressions with phase shifts The compound-angle formulas can be used to epand trigonometric epressions. Epand cos θ +. WritE State an appropriate identit. cos(a + B) = cos(a)cos(b) sin(a)sin(b) Let A = θ and B =. Substitute for eact values. Since cos cos θ + = cos(θ)cos sin(θ)sin = and sin =, cos θ + = cos(θ) sin(θ) Simplif. = cos(θ) sin(θ) State the answer. cos θ + WorKeD example 7 simplifi cation of sin = cos(θ) sin(θ) n ± θ and cos n ± θ for n Z Recall that cos θ = sin(θ) and sin θ = cos(θ) as complementar angles. Compound-angle formulas can be used to simplif and verif man of these results and similar formulas from earlier results, that is trigonometric epansions of the forms n sin ± θ and cos n ± θ where n Z. Use compound-angle formulas to simplif cos θ. = WritE State an appropriate identit. cos(a B) = cos(a)cos(b) + sin(a)sin(b) Let A = and B = θ. cos θ = cos cos (θ) + sin sin (θ) 7 MaTHs QuesT specialist MaTHeMaTICs VCe units and

18 Simplif and use eact values. Since cos cos State the final answer. cos WorKeD example 8 = 0 and sin eact values for multiples of Eact values are known for the trigonometric ratios for all multiples of radians or 0, and for all multiples radians or 5. Using the compound-angle formulas the eact value can be found for a trigonometric ratio of an angle that is an odd multiple of radians or 5. This can be obtained b rewriting the multiple of radians or 5 as a sum or difference of known fractions in terms of multiples of radians or 0 and radians or 5. Find the eact value of sin. =, θ = 0 cos (θ) + sin (θ) θ = sin (θ) WritE 5 Rewrite the argument as a sum or difference of fractions. + =, or in degrees, = sin = sin + State an appropriate identit. sin(a + B) = sin(a)cos(b) + cos(a)sin(b) Let A = 5 and B =. sin = sin 5 cos + cos 5 sin Simplif and use eact values. Substitute sin 5 =, cos =, cos 5 = and sin =. sin = + = Simplif and state the final answer. sin = ( ) Topic TrIgonoMeTr 7

19 WorKeD example 9 using triangles to fi nd values B drawing triangles to find the values of trigonometric ratios of a single angle and then using the compound-angle formulas, the trigonometric values of the addition or subtraction of two angles ma be found. If cos(a ) = 7 and sin(b) = 5, where 0 < A < and < B <, find the eact value of sin(a B). State the values of the sides of the required right-angled triangle. Use Pthagoras theorem to calculate the third side length. State the third side length of the triangle. Draw the triangle. State the value of the unknown trigonometric ratio. 5 State the values of the sides of another required right-angled triangle. Use Pthagoras theorem to calculate the third side length. 7 State the third side length of the triangle. Draw the triangle. 8 Calculate the value of the unknown trigonometric ratio. WritE/draW cos(a) = = adjacent hpotenuse The adjacent side length is and the hpotenuse is. = 9 = 5 = 5 The other side length is 5. We know that 5,, is a Pthagorean triad. Given that 0 < A <, so A is in the first quadrant, sin (A) = 5. sin (B) = 7 5 = opposite hpotenuse The opposite side length is 7 and the hpotenuse is = 5 9 = 57 = The other side length is. We know that 7,, 5 is a Pthagorean triad. Since < B <, B is in the second quadrant, B is an obtuse angle and cosine is negative in the second quadrant. Therefore, cos(b) = 7 5. A 5 B 5 7 MaTHs QuesT specialist MaTHeMaTICs VCe units and

20 9 State and use an appropriate identit. sin(a B) = sin(a)cos(b) cos(a)sin(b) 0 Substitute for the values and simplif. sin(a B) = Simplif and state the final answer. sin(a B) = 0 Eercise. PRactise Consolidate Compound-angle formulas WE5 Evaluate sin(5 )cos(9 ) + cos(5 )sin(9 ). Find the value of cos(7 )cos( ) sin(7 )sin( ). WE Epand cos θ +. Epress sin θ + as a combination of sines and cosines. 5 WE7 Use compound-angle formulas to simplif cos( θ). Simplif sin( θ). 7 WE8 Find the eact value of sin 8 Find the eact value of tan WE9 If cos(a) = 8 and sin(b) = 5 7 where 0 < A < and < B <, find the eact value of sin(a B). 0 Given that tan(a) = 9 7 and cos(b) = 0 5 where < A < and 0 < B <, find the eact value of cos(a + B). Evaluate each of the following. a sin(7 )cos( ) + cos(7 )sin( ) b cos(7 )cos( ) sin(7 )sin( ) c cos(7 )cos( ) + sin(7 )sin( ) d cos( )sin(8 ) sin( )cos(8 ) Evaluate each of the following. tan(5 ) tan( ) a + tan(5 )tan( ) Epand each of the following. a sin θ c cos θ b tan( ) + tan(8 ) tan( )tan(8 ) b sin θ + 5 d cos θ + Topic Trigonometr 75

21 Master.5 Use compound-angle formulas to simplif each of the following. a sin θ b cos θ c sin( + θ) d cos( θ) 5 Use compound-angle formulas to simplif each of the following. a sin θ b cos + θ c tan( θ) d tan( + θ) Simplif each of the following. a sin + sin c cos b tan + tan + cos d cos cos + 7 Find each of the following in eact simplest surd form. a cos 7 b tan c sin d tan 8 Given that cos(a) =, sin(b) =, and A and B are both acute angles, find the 5 eact value of: a cos(a B) b tan(a + B). 9 Given that sin(a) = 5 and tan(b) = where A is obtuse and B is acute, find the 7 eact value of: a sin(a + B) b cos (A + B). 0 Given that sec(a) = 7, cosec(b) =, and A is acute but B is obtuse, find the eact value of: a cos(a + B) b sin(a B). Given that cosec(a) = a, sec(b) =, A and B are both acute, and 0 < a < and b 0 < b <, evaluate tan(a + B). Given that sin(a) = a and cos(b) = a where A and B are both acute, a + a + evaluate tan(a + B). Double-angle formulas In this section we consider the special cases of the addition formulas when B = A. Double-angle formulas In the formula sin(a + B) = sin(a)cos(b) + cos(a)sin(b), let B = A. sin(a) = sin(a)cos(a) + sin(a)cos(a) sin(a) = sin(a)cos(a) In the formula cos(a + B) = cos(a)cos(b) sin(a)sin(b), let B = A. cos(a) = cos(a)cos(a) sin(a)sin(a) cos(a) = cos (A) sin (A) 5 7 Maths Quest SPECIALIST MATHEMATICS VCE Units and

22 WorKeD example 0 Since sin (A) + cos (A) =, it follows that cos (A) = sin (A). This formula can be rewritten in terms of sin(a) onl. cos(a) = ( sin (A)) sin (A) cos(a) = sin (A) Alternativel, if we substitute sin (A) = cos (A), then this formula can also be rewritten in terms of cos(a) onl. cos(a) = cos (A) ( cos (A)) cos(a) = cos (A) There are thus three equivalent forms of the double-angle formulas for cos(a). tan(a) + tan(b) If we let B = A in the formula tan(a + B) =, we obtain tan(a)tan(b) tan(a) + tan(a) tan(a) = tan(a)tan(a) tan(a) = tan(a) tan (A) All of these formulas can be summarised as follows: sin(a) = sin(a)cos(a) cos(a) = cos (A) sin (A) cos(a) = sin (A) cos(a) = cos (A) tan(a) = tan(a) tan (A) using double-angle formulas in simplifing epressions The double-angle formulas can be used to simplif man trigonometric epressions and can be used both was; for eample, sin(a)cos(a) = sin(a). Find the eact value of sin 7 7 cos. WritE State an appropriate identit. sin(a)cos(a) = sin(a) Let A = 7. 7 sin cos 7 = 7 sin Simplif. Since 7 = 7, sin 7 cos 7 = sin 7 Topic TrIgonoMeTr 77

23 Use the eact values to substitute into the epression. State the answer. sin WorKeD example If cos(a ) =, determine the eact values of: a sin (A ) b cos (A ) c tan (A ). State the values of the sides of the required right-angled triangle. Draw the triangle and label the side lengths using the definition of the trigonometric ratio. Label the unknown side length as. Use Pthagoras theorem to calculate the third unknown side length. Redraw the triangle. Finding trigonometric epressions involving double-angle formulas We can use the double-angle formulas to obtain eact values for trigonometric epressions. 5 Appl the definitions of the sine and tangent functions. WritE/draW cos(a) = = adjacent hpotenuse The adjacent side length is and the hpotenuse is. A + = = = 5 5 A 7 Substitute sin sin 7 = : = 7 cos 7 = cos(a) = 5, sin(a) = and tan(a) = 5 78 MaTHs QuesT specialist MaTHeMaTICs VCe units and

24 a Use the required identit. Substitute the known values and simplif. b Using the required identit, choose an one of the three choices for cos(a). Substitute the known values and simplif. c State the required identit. Substitute the known values and simplif the ratio. As an alternative method, use the double-angle formulas for tan. Substitute for the known value and simplif. WorKeD example a sin(a) = sin(a)cos(a) = 5 = 5 8 b cos(a) = cos (A) sin (A) cos(a) = 5 = 5 = 7 8 c tan(a) = sin(a) cos(a) = = 5 tan(a) = tan(a) tan (A) tan(a) = 5 tan(a) = tan(a) tan (A) 7 = 5 ( 5) = 5 = 5 7 solving trigonometric equations involving double-angle formulas Trigonometric equations are often solved over a given domain, usuall [0, ]. In this section we consider solving trigonometric equations that involve using the double-angle formulas. Solve for if sin() + Epand and write the equation in terms of one argument onl. cos() = 0 for [0, ]. WritE Use sin() = sin()cos() sin() + cos() = 0 sin()cos() + cos() = 0 Topic TrIgonoMeTr 79

25 Factorise b taking out the common factor. cos()( sin() + ) = 0 Use the Null Factor Law. cos() = 0 or sin() + = 0 sin() = Solve the first equation. cos() = 0 =, 5 Solve the second equation. sin() = WorKeD example Trigonometric identities using double-angle formulas Previousl we used the fundamental trigonometric relationships to prove trigonometric identities using the reciprocal trigonometric functions. In this section we use the compound-angle formulas and the double-angle formulas to prove more trigonometric identities. Prove the identit cos(a )cos(a ) + sin(a )sin(a ) sin(a)cos(a( ) cos(a)sin(a( ) = cosec(a ). WritE cos(a)cos(a) + sin(a)sin(a) Start with the left-hand side. LHS = sin(a)cos(a) cos(a)sin(a) Simplif the numerator and denominator b recognising these as epansions of appropriate compound-angle identities. cos(a A) = sin(a A) Simplif. = cos(a) sin(a) cos(a) Epand the denominator using the = double-angle formula. sin(a)cos(a) 5 Simplif b cancelling the common factor. The proof is complete. = sin(a) Since sin(a) = cosec(a), LHS = sin(a) = cosec(a) = RHS as required. =, 5 State all solutions of the original equation. =,,, 5 80 MaTHs QuesT specialist MaTHeMaTICs VCe units and

26 WorKeD example Half-angle formulas If we replace A with A, the double-angle formulas can be written as the half-angle formulas. sin(a) = sin A cos A cos(a) = cos A = cos A sin = sin A These can also be rearranged and are often used as: cos(a) = sin A + cos(a) = cos A Prove the identit cosec(a ) cot(a ) = tan A. WritE Start with the left-hand side. LHS = cosec(a) cot(a) Use cosec(a) = cos(a) and cot(a) = sin(a) sin(a). = sin(a) cos(a) sin(a) Form the common denominator. = cos(a) sin(a) Use appropriate half-angle formulas. = 5 Simplif b cancelling the common factors. The proof is complete. A sin A sin A cos A sin A = cos A = tan A = RHS Topic TrIgonoMeTr 8

27 WorKeD example 5 Multiple-angle formulas There are man other trigonometric formulas for multiple angles. For eample: sin(a) = sin(a) sin (A) cos(a) = cos (A) cos(a) tan(a) = tan(a) tan (A) tan (A) sin(a) = cos(a)( sin(a) 8 sin (A)) cos(a) = 8 cos (A) 8 cos (A) + tan(a) = tan(a)( tan (A)) tan (A) + tan (A) Some of these proofs are provided in the net worked eample; the remaining ones are left for the eercises. Prove the identit cos(a) = cos (A ) cos(a ). WritE Start with the left-hand side. LHS = cos(a) Epand the multiple argument. cos(a + A) = cos(a)cos(a) sin(a)sin(a) Since we want the right-hand side in terms of cosines, substitute using appropriate trigonometric identities. cos(a) = cos (A) sin(a) = sin(a)cos(a) cos(a) = cos(a)( cos (A) ) sin(a)( sin(a)cos(a)) Epand the brackets. cos(a) = cos (A) cos(a) sin (A)cos(A) 5 Rearrange the epression in terms of cosines. Epand the brackets and simplif. The proof is complete. EErcisE.5 PractisE Double-angle formulas Substitute sin (A) = cos (A): cos(a) = cos (A) cos(a) ( cos (A))cos(A) cos(a) = cos (A) cos(a) (cos(a) cos (A)) = cos (A) cos(a) = RHS 5 WE0 Find the eact value of sin 8 cos 5 8. Find the eact value of cos (57 0 ). WE If cos(a) =, determine the eact values of: a sin(a) b cos(a) c tan(a). Given that tan(a) =, determine the eact values of: 7 a sin(a) b cos(a) c tan(a). 5 WE Solve for if sin() cos() = 0 for [0, ]. 8 MaTHs QuesT specialist MaTHeMaTICs VCe units and

28 Consolidate Find the values of [0, ] if sin() + cos() = 0. 7 WE Prove the identit 8 Prove the identit tan sin(a)cos(a) cos(a)sin(a) cos(a)cos(a) + sin(a)sin(a) = sin(a). + A + tan A = sec(a). 9 WE Prove the identit cosec(a) + cot(a) = cot 0 Prove the identit cos(a) + cos(a) = tan A. A. WE5 Prove the identit sin(a) = sin(a) sin (A). Prove the identit tan(a) = tan(a) tan (A). tan (A) Evaluate each of the following epressions, giving our answers in eact form. a sin 8 cos 8 c sin ( 0 ) b cos ( 0 ) sin ( 0 ) d tan tan Given that sec(a) = 8, find the eact values of: a sin(a) b cos(a) c tan(a). 5 Solve each of the following equations for [0, ]. a sin() = sin() b cos() = cos() c sin() = cos() d sin() = cos() Solve each of the following equations for [0, ]. a tan() = sin() b sin() = cos() c sin() = sin() d cos() = sin() For questions 7, prove each of the given identities. 7 a b 8 a sin(a)cos(a) cos(a)sin(a) cos(a)cos(a) + sin(a)sin(a) = tan(a) cos(a)cos(a) + sin(a)sin(a) sin(a)cos (A) cos(a)sin(a) = cot(a) c sin(a) sin(a) cos(a) cos(a) = d cos(a) sin(a) + sin(a) cos(a) = cot(a) b tan(a) + tan(a) tan(a) tan(a) = cos(a) tan(a) tan(b) sin(a B) = tan(a) + tan(b) sin(a + B) Topic Trigonometr 8

29 Master c d 9 a c sin(a) cos(a) sin(a) + cos(a) sin(a) + cos(a) sin(a) cos(a) = tan(a) cos(a) + sin(a) cos(a) sin(a) + cos(a) sin(a) cos(a) + sin(a) = sec(a) sin(a) cos(a) = cot A b sin(a) + cos(a) = tan A cos(a) + sin(a) + cos(a) + sin(a) = tan(a) d sin(a) + sin(a) + cos(a) + cos(a) = tan(a) 0 a sin(a + B)sin(A B) = sin (A) sin (B) b tan(a + B)tan(A B) = tan (A) tan (B) tan (A)tan (B) c cot(a + B) = cot(a)cot(b) cot(a) + cot(b) d cot(a B) = cot(a)cot(b) + cot(b) cot(a) a sin(a) = tan(a) + tan (A) b cos(a) = tan (A) + tan (A) c cos (A) sin (A) = + cos(a) sin(a) sin(a) d cos (A) + sin (A) = cos(a) + sin(a) sin(a) In a triangle with side lengths a, b and c, where C is a right angle and c the hpotenuse, show that: a sin(a) = ab b cos(a) = b a c tan(a) = ab c c b a d sin A = c b c e cos A = c + b c Chebshev (8 89) was a famous Russian mathematician. Although he is known more for his work in the fields of probabilit, statistics, number theor and differential equations, Chebshev also devised recurrence relations for trigonometric multiple angles. One such result is cos(n) = cos()cos((n )) cos((n )). Using this result, show that: a cos(a) = 8 cos (A) 8 cos (A) + b cos(5a) = cos 5 (A) 0 cos (A) + 5 cos(a) c cos(a) = cos (A) 8 cos (A) + 8 cos (A). f tan A = c b c + b. 8 Maths Quest SPECIALIST MATHEMATICS VCE Units and

30 . Units & AOS Topic Concept 5 Restricted circular functions Concept summar Practice questions Chebshev s recurrence formula for multiple angles of the sine function is sin(n) = cos()sin((n )) sin((n )). Using this result, show that: a sin(a) = cos(a)( sin(a) 8 sin (A)) b sin(5a) = sin 5 (A) 0 sin (A) + 5 sin(a) c sin(a) = cos(a)( sin 5 (A) sin (A) + sin(a)). Inverse trigonometric functions Inverse functions All circular functions are periodic and are man-to-one functions. Therefore, the inverses of these functions cannot be functions. However, if the domain is restricted so that the circular functions are one-to-one functions, then their inverses are functions. The inverse sine function The sine function, = sin(), is a man-to-one function. 0 Therefore, its inverse does not eist as a function. However there are man restrictions of the domain, such as,,, or,, that will ensure it is a one-toone function. For convenience, let, restricted sine function. f :, [, ] (, ) 0 be the domain and [, ] the range of the (, ) Therefore, it is a one-to-one function and its inverse eists. The inverse of this function is denoted b sin. (An alternative notation is arcsin.) Topic Trigonometr 85

31 WorKeD example The graph of = sin () is obtained from the graph of = sin() b reflection in the line =. f : [, ] (, ),, f () = sin () 0, ( ) There are an infinite number of solutions to sin() =, for eample, + and +, since we can alwas add an multiple of to an angle and get the same result. However, sin means sin() = and,, so there is onl one solution in this case:. Find each of the following. a sin () b sin sin 5 WritE a Write an equivalent statement. a = sin () sin() = c sin(sin (0.5)) State the result. This does not eist. There is no solution to sin() =. b Use the known results. Write an equivalent statement and state the result. b Since sin = sin 5 sin =, 5 sin() = and, = sin The onl solution is =. sin 5 sin = c State the result. c sin(sin (0.5)) = MaTHs QuesT specialist MaTHeMaTICs VCe units and

32 General results for the inverse sine function In general, we have the following results for the inverse sine function: f : [, ],, f () = sin () sin(sin ()) = if [, ] The inverse cosine function sin (sin()) = if, The cosine function, = cos(), is a man-to-one function. 0 Therefore, its inverse does not eist as a function. However, there are man restrictions of the domain, such as [, 0] or [0, ] or [, ], that will ensure it is a one-to-one function. Let [0, ] be the domain and [, ] the range of the restricted cosine function. f : [0, ] [, ] where f() = cos(). 0 (0, ) (, ) Therefore, it is a one-to-one function and its inverse eists. The inverse of this function is denoted b cos. (An alternative notation is arccos.) Topic Trigonometr 87

33 WorKeD example 7 The graph of = cos () is obtained from the graph of = cos() b reflection in the line =. f : [, ] [0, ], f() = cos () (, ) 0 0, ( ) (, 0) There are an infinite number of solutions to cos() =, for eample, +, +, and, since we can alwas add an multiple of to an angle. However, cos means cos() = and [0, ], so there is onl one solution, namel. Find each of the following. a cos b cos WritE a Write an equivalent statement. a = cos cos 5 c cos cos State the result. This does not eist. There is no solution to cos() =. b Use the known results. Write an equivalent statement and state the result. b Since cos = cos 5 cos =, 5 cos() = and [0, ] The onl solution is =. cos cos 5 c State the result. c cos cos = = = cos 88 MaTHs QuesT specialist MaTHeMaTICs VCe units and

34 WorKeD example 8 general results for the inverse cosine function In general, we find that: f : [, ] [0, ], f () = cos () cos(cos ()) = if [, ] cos (cos()) = if [0, ] Find the eact value of cos sin The inverse trigonometric functions are angles. Draw a right-angled triangle and label the side lengths using the definition of the trigonometric ratios. Calculate the value of the third side using Pthagoras.. WritE/draW Let θ = sin so that sin(θ) =. θ + = + = 9 = 9 = 8 = State the required value. cos sin θ = cos(θ) = Double-angle formulas Sometimes we ma need to use the double-angle formulas. sin(a) = sin(a)cos(a) cos(a) = cos (A) sin (A) = cos (A) = sin (A) Topic TrIgonoMeTr 89

35 WorKeD example 9 Find the eact value of sin cos 5. WritE/draW The inverse trigonometric functions are angles. Let θ = cos so that cos(θ) =. 5 5 Draw a right-angled triangle and label the side lengths using the definition of the trigonometric ratios. Calculate the value of the third side using Pthagoras theorem. Use an appropriate double-angle formula. The inverse tangent function The tangent function, = tan(), is a man-to-one function. 0 = + = 5 + = 5 = 5 = = θ = = = Therefore, its inverse does not eist as a function. However, there are man restrictions of the domain, such as, or, or,, that will ensure it is a one-to-one function. 5 sin(θ) = sin(θ)cos(θ) = State what is required. sin cos = 5 90 MaTHs QuesT specialist MaTHeMaTICs VCe units and

36 Let, be the domain and R the range of the restricted tangent function. Note that we must have an open interval, because the function is not defined at = ± ; at these points we have vertical asmptotes. f :, R, f () = tan() 0 = = Therefore, it is a one-to-one function and its inverse eists. The inverse of this function is denoted b tan. (An alternative notation is arctan.) The graph of = tan () is obtained from the graph of = tan() b reflection in the line =. f : R, where f () = tan () Note that there horizontal asmptotes at = ±. 0 = = There are an infinite number of solutions to tan() =, for eample,, + and +, since we can alwas add an multiple of to an angle. However, tan ( ) means tan() = and,, so there is onl one solution:. Topic Trigonometr 9

37 WorKeD example 0 Find: a tan a Use the known results. tan Write an equivalent statement and state the result. general results for the inverse tan function In general, we find that: f : R WritE a tan b tan(tan ()). Let tan = tan,, f () = tan () tan(tan ()) = if R = tan tan() = and, The onl solution is =. tan tan b State the result. b tan(tan ()) = WorKeD example = tan (tan()) = if, Double-angle formulas It ma be necessar to use the double-angle formulas, such as tan(a) = tan(a) tan (A). Find the eact value of tan tan WritE The inverse trigonometric functions are angles. Let θ = tan so that tan(θ) =. Use the double-angle formulas. tan(θ) = tan(θ) tan (θ). = = = State the result. tan tan = = 9 MaTHs QuesT specialist MaTHeMaTICs VCe units and

38 WorKeD example Compound-angle formulas We ma also need to use the compound-angle formulas: sin(a + B) = sin(a)cos(b) + cos(a)sin(b) sin(a B) = sin(a)cos(b) cos(a)sin(b) cos(a + B) = cos(a)cos(b) sin(a)sin(b) cos(a B) = cos(a)cos(b) + sin(a)sin(b) Evaluate cos sin The inverse trigonometric functions are angles. Use the definitions of the inverse trigonometric functions. Draw the right-angled triangle and state the unknown side length using well-known Pthagorean triads. tan(a) + tan(b) tan(a + B) = tan(a)tan(b) tan(a) tan(b) tan(a B) = + tan(a)tan(b) tan. WritE/draW Let A = sin and B = tan Thus, sin(a) = and tan(b) =. A 5 State the ratios from the triangles. sin(a) =, cos(a) = 5 Substitute the ratios into the compound-angle formulas. 5 State the required result. cos sin 5 B. sin(b) =, cos(b) = 5 5 cos(a B) = cos(a)cos(b) + sin(a)sin(b) = tan = 5 5 Topic TrIgonoMeTr 9

39 WorKeD example Determining maimal domains and ranges For = sin (), the domain is [, ] and the range is,. For = cos (), the domain is [, ] and the range is [0, ]. For = tan (), the domain is R and the range is,. For inverse trigonometric functions that have been dilated or translated, we can appl these dilations and translations to determine the domain and range of the transformed function. State the domain and range of: a = cos 5 b = tan 7 WritE a = cos () has a domain of [, ]. a 5 Use the definition of the modulus function. 5 Solve the inequalit State the domain. = cos 5 7 or, has a maimal domain of 5 = cos () has a range of [0, ]. There is a dilation b a factor of parallel to the -ais and a translation of units down parallel to the -ais. The range is from 0 to. State the range. = cos 5 b = tan () has a domain of R. b = tan 7 = tan () has a range of, State the range. = tan 7 ( +, + ). has a range of [, ]. + has a domain of R.. There is a dilation b a factor of parallel to the -ais and a translation of unit up parallel to the -ais. The range is from + to +, not including the end points. + has a range of 9 MaTHs QuesT specialist MaTHeMaTICs VCe units and

40 Eercise. Inverse trigonometric functions PRactise Consolidate WE Find each of the following. a sin (.) b sin sin Find each of the following. a sin 5 b sin WE7 Find each of the following. sin a cos (.) b cos cos Find each of the following. a cos b cos cos 5 WE8 Find the eact value of cos sin Find the eact value of sin cos 7 WE9 Find the eact value of sin cos 8 Find the eact value of cos sin 9 WE0 Find: a tan 0 Find: a tan tan tan 7 5 WE Find the eact value of tan tan Find the eact value of cot tan WE Evaluate sin cos Evaluate tan sin 5 5 cot tan 5 5 WE State the domain and range of: b tan (tan (.)). b tan tan 5 c sin (sin (0.9)) c sin sin c cos cos c cos cos a = sin 5 b = 5 tan State the domain and range of: a = cos ( + 5) b = 8 tan (0) +. 7 Evaluate each of the following. a sin () b sin (.) c sin d cos ( ) e cos g tan ( ) h tan. +. f cos (.) Topic Trigonometr 95

41 8 Evaluate each of the following. a sin (sin(.)) b sin sin d cos (cos(0.5)) e cos cos g tan tan 8 h tan tan Evaluate each of the following. a sin cos b cos sin d sin (tan ( )) e cos tan 0 Evaluate each of the following. a sin cos b tan cos 9 d sin tan 5 8 e cos sin Evaluate each of the following. a sin cos b tan sin d sin tan Evaluate each of the following. a sin cos + sin c cos tan 5 Show that: cos 9 a cos tan = c sin tan 5 = e tan () tan 5 = Show that: a sin = 5 sin 5 c cos = cos 7 8 e tan = tan 5 e tan cos 5 b cos cos d sin tan c sin f cos sin cos 5 c tan sin 5 f tan cos c tan sin 5 f cos tan 7 c cos tan f cos sin sin 5 sin 0 b sin + 5 tan = d tan () tan = f tan (5) tan =. b sin 7 = 5 sin 5 d cos = cos 9 f tan = tan 8 5 State the implied domain and range of each of the following. a = sin ( ) b = cos ( ) c = tan ( ) d = 5 sin e = cos 5. f = 7 tan Maths Quest SPECIALIST MATHEMATICS VCE Units and

42 Master.7 State the implied domain and range of each of the following. a = sin ( ) + c = 5 tan ( + ) e = 5 cos 7 b = cos ( 5) d = sin 5 f = 8 5 tan 7 a State a sequence of transformations that, when applied to = sin (), produce the graph of = a + b sin. Hence, state the domain and c range of = a + b sin c. b State a sequence of transformations that, when applied to = cos (), produce the graph of = a + b cos (c). Hence, state the domain and range of = a + b cos (c). c State a sequence of transformations that, when applied to = tan (), produce the graph of = a + b tan. Hence, state the domain and c range of = a + b tan c. 8 Show that: a sin () = cos ( ) for [0, ] b tan () + tan = for > 0 c cos () = tan d sin e cos for (0, ) a + tan b a + b a = for a > 0 and b > 0 a + tan a a + b b = for a > 0 and b > 0 f tan () tan + = for > g sin + = cos + = tan for > 0. General solutions of trigonometric equations In this section consideration is given to the general solutions of trigonometric equations, rather than finding the solutions over a specified domain. Trigonometric equations can have an infinite number of solutions. To epress the possible solutions mathematicall, we derive formulas that will give the general solution in terms of an natural number n, where n Z. + + Topic Trigonometr 97

43 WorKeD example general solutions involving cosines Consider the equation cos() = a. One answer is = cos (a). If 0 < a <, then 0 < <, so is in the first quadrant. θ Because cosine is positive in the first and fourth quadrant, there is also another answer, = cos θ (a). We can add or subtract an multiple of to either answer and obtain an equivalent angle. cos() = a = cos (a), + cos (a), + cos (a), = cos (a), cos (a), cos (a), The totalit of solutions can be represented as = n ± cos (a), where n Z. Although we have demonstrated this result for 0 < a <, it is in fact true for a. The general solution of cos( ) = a where a The general solution of cos() = a where a is given b = n ± cos (a), where n Z. Find the general solution to the equation cos () =. WritE State one solution. = cos = State the general solution. = n ± Take out a common factor so that the general solution can be written in simplest form. = (n ± ) where n Z general solutions involving sines Consider the equation sin() = a. One answer is = sin (a), and if 0 < a <, then 0 < < θ, so is in the first quadrant. Since sine is positive in the first and second quadrants, there is also another answer, = sin (a). We can add or subtract an multiple of to either answer and obtain an equivalent angle. sin() = a = sin (a), + sin (a), + sin (a), = sin (a), sin (a), 5 sin (a), If n is an integer, then n is an even integer and n + is an odd integer. θ 98 MaTHs QuesT specialist MaTHeMaTICs VCe units and

44 WorKeD example 5 The totalit of solutions can be represented as = n + sin (a) or = (n + ) sin (a), where n Z. Although we have demonstrated this result for 0 < a <, it is true for a. The general solution of sin( ) = a where a The general solution of sin() = a where a is given b = n + sin (a), (n + ) sin (a), where n Z. Find the general solution to the equation sin() =. WritE State one solution. = sin State the general solution. = n + or = (n + ) Take out common factors in the first solution so that the general solution can be written in simplest form. Take out common factors in the second solution = n + = (n + ) general solutions involving tangents Consider the equation tan() = a. One answer is = tan (a), and if a > 0, then 0 < <, so is in the first quadrant. Since θ tangent is positive in the first and third quadrants, there is also another answer, = + tan (a). + θ We can add or subtract an multiple of to either answer and obtain an equivalent angle. tan() = a = tan (a), + tan (a), + tan (a), = + tan (a), + tan (a), 5 + tan (a), The totalit of solutions can be represented as one solution: = n + tan (a), where n Z. = = (n + ) = n + = n + = (n + ) 5 State the general solution. = (n + ), (n + ) where n Z Topic TrIgonoMeTr 99

45 WorKeD example The general solution of tan( ) = a The general solution of tan() = a where a R is given b = n + tan (a), where n Z. Although we have demonstrated this result onl for a > 0, it is true for a R. Find the general solution to the equation tan() =. WritE State one solution. = tan ( ) = State the general solution. = n + Take out a common factor so that the general solution can be written in simplest form. WorKeD example 7 = (n + ) where n Z When solving more complicated trigonometric equations, often multiple solutions eist. We ma be required to find all solutions to each part of the equation being considered. Find the general solution to the equation cos () = 0. Make the trigonometric function the subject. Use the formula to find the general solution of the first equation. WritE cos () = 0 So that: cos () = cos() = ± () cos() = or () cos() = cos() = = n ± cos = n ± = (n ± ) = (n ± ) 00 MaTHs QuesT specialist MaTHeMaTICs VCe units and

46 Use the formula to find the general solution of the second equation. cos() = = n ± cos = n ± 5 = (n ± 5) = (n ± 5) State the final general solutions. = (n ± ) or = (n ± 5) where n Z WorKeD example 8 general solutions involving phase shifts When solving trigonometric equations involving phase shifts, we must solve the resulting equations for the unknown values of. Find the general solution of sin + + = 0. WritE Make the trigonometric function the subject. sin + + = 0 sin + = sin + = Use the formula to state the general solution. () + = n + sin or () + = (n + ) sin Solve the first equation. + = n + sin + = n = n = (n ) = (n ) Topic TrIgonoMeTr 0

47 Solve the second equation. + = (n + ) sin + = n + + = (n + ) = (n + ) 5 State the final solutions. = (n ) or (n + ) n Z WorKeD example 9 equations reducible to quadratics Equations can often be reduced to quadratics under a suitable substitution. Find the general solution of the equation sin () + sin() = 0. WritE Use a substitution. sin () + sin() = 0 Let u = sin(). Factorise. u + u = 0 (u )(u + ) = 0 Substitute back for u. ( sin() )(sin() + ) = 0 Solve the trigonometric equation using the Null Factor Law. 5 Find the general solution of the first equation. () sin() = or sin() = = n + sin = n + = (n + ) = (n + ) sin() = = (n + ) sin = n + = n + 5 = (n + 5) = (n + 5) () sin() = 0 MaTHs QuesT specialist MaTHeMaTICs VCe units and

48 Find the general solution of the second equation. sin() = = n + sin ( ) 7 Sometimes some parts of the solution are alread included in some other parts. Give n some values. 8 State all the general solutions of the equation. WorKeD example 0 = n = (n ) = (n ) sin() = = (n + ) sin ( ) = n + + = n + = (n + ) = (n + ) Let n = 0,,,,. = (n ) =,, 7,, 5 = (n + ) =, 7,, 5 We can see that the solution = (n + ) incorporates all the solutions from = (n ). = (n ) or (n + ) or (n + 5) where n Z. Trigonometric equations involving multiple angles We can find the general solutions to trigonometric equations involving multiple angles b appling the general solution formulas rather than epanding the multiple angles. Find the general solution to cos() = sin(). Rewrite using one trigonometric function. Convert sines into cosines, since the solution for cosine is easier to work with. WritE Use sin A cos() = sin() = cos(a). cos() = cos Topic TrIgonoMeTr 0

49 Solve using an appropriate general solution. = n ± Comparison of eamples Note that the last two worked eamples, 9 and 0, are in fact the same, as cos() = sin() cos(()) = sin() b double-angle formulas (sin()) = sin() sin () + sin() = 0 () = n + or () = n Solve the first equation. = n + = n + = (n + ) = (n + ) Solve the second equation. = n + = n = (n ) = (n ) 5 State the general solutions of the equation. = (n + ) or (n ) where n Z. and therefore the should have the same general solution. The two given answers do not appear to be the same, although one answer, = (n ), is common to both. This situation is ver common in these tpes of problems. However, if we substitute values of n, the two results generate the same particular solutions. When n = 0,,,,, 5, from Worked eample 9: = (n ) =,, 7,, 5, 9, = (n + ) =,, 5 = (n + 5) 5 =, 7 0 Maths Quest SPECIALIST MATHEMATICS VCE Units and

50 Eercise.7 PRactise Consolidate When n = 0,,,,, 5, from Worked eample 0: = (n ) =,, 7,, 5, 9, = (n + ) =, 5, 9,, 7,, 5 It is interesting to compare these results to those obtained b CAS calculators. In some cases a calculator will not solve the equation for the general solution, and in other cases it will. The solution obtained b CAS ma be in a different form to our answers above. The results ma be given differentl depending on the MODE, which could be set to either Eact or Auto. General solutions of trigonometric equations WE Find the general solution to the equation cos() = Determine the general solution of cos() + = 0. WE5 Find the general solution to the equation sin() =. Determine the general solution of sin() + = 0. 5 WE Find the general solution to the equation tan() =. Find the general solution to tan() + = 0. 7 WE7 Find the general solution to the equation cos () = 0. 8 Find the general solution to the equation tan () = 0. 9 WE8 Find the general solution of sin + = 0. 0 Find the general solution of cos + = 0. WE9 Find the general solution to sin () sin() + = 0. Find the general solution to the equation cos () + cos() = 0. WE0 Find the general solution to cos() = sin(). Find the general solution to cos() = sin(). 5 Find the general solution to each of the following equations. a cos() = 0 b cos() + = 0 c sin() + = 0 d sin() + = 0 Find the general solution to each of the following equations. a sin () = 0 b sin () = 0 c cos () = 0 d cos () = 0 7 Find the general solution to each of the following equations. a tan() + = 0 b tan() = 0 c tan () = 0 d tan () = 0 8 Find the general solution to each of the following equations. a sin () + sin() = 0 b cos () cos() = 0 c cos () + cos() = 0 d sin () sin() = 0. Topic Trigonometr 05

51 Master.8 Units & AOS Topic Concept Sketch graphs of reciprocal circular functions Concept summar Practice questions 9 Find the general solution to each of the following equations. a sin () + sin() + = 0 b cos () cos() + = 0 0 Find the general solution to each of the following equations. a sin = 0 b sin + Find the general solution to each of the following equations. a cos + + = 0 b cos Find the general solution to each of the following equations. a tan + = 0 b tan Find the general solution to each of the following equations. a tan () + ( + )tan() + = 0 b tan () + ( )tan() = 0 Find the general solution to each of the following equations. a sin() = sin() b cos() = cos() 5 Find the general solution to each of the following equations. a sin () + sin () sin() = 0 b cos () cos () cos() + = 0 Find the general solution to each of the following equations. a tan () tan () tan() + = 0 b tan () tan () + = 0 + = 0 = 0 + = 0 Graphs of reciprocal trigonometric functions Topic described how the graph of f() can be used to find the graph of. We can f() use this method to graph sec() = cos(), cosec() = and cot() = sin() tan(). The graph of = sec( ) Consider the graph of = cos(). 0 0 Maths Quest SPECIALIST MATHEMATICS VCE Units and

52 In the portion of the graph shown, the -intercepts occur at,, and. This means that the reciprocal function will have vertical asmptotes at =, =, = and =. The horizontal asmptote will be = 0. The graph of = cos() is below the -ais for < < and passes through the point (, ). This means that = will also be below the -ais in this cos() interval and will pass through the point (, ). It will follow a similar pattern in the region < <. In the region < <, the graph of = cos() is above the -ais and passes through the point (0, ). This means that = will also be above the -ais and cos() will pass through (0, ). The graph of = (or = sec()) is shown below. cos() = 0 = = = The graph of = cosec( ) In a similar fashion, the graph of = sin() can be used to determine the graph of = (or = cosec()). sin() The graph of = sin() is shown below. Note that in this instance, the -intercepts occur at,, 0, and. 0 Topic Trigonometr 07

53 WorKeD example Sketch = cos(). Period: Amplitude: Horizontal shift: 0 Vertical shift: 0 Find the -intercepts and hence the vertical asmptotes for the reciprocal graph. The graph of = sin() = looks like the following. = 0 = 0 = = Use the graph of = cos() to sketch = over the domain cos(). WritE/draW (, ) (, ) 0 -intercepts occur at =, =, = and =. These will be the vertical asmptotes for the reciprocal function. (, ) (, ) 0 = = = = 08 MaTHs QuesT specialist MaTHeMaTICs VCe units and

54 The graph of = cos() is above the -ais in the regions <, < < and <. The graph of = will also be cos() above the -ais in these regions. A maimum value of = is reached in the original graph, meaning that a minimum of = will be reached in the reciprocal graph. The graph of = cos() is below the -ais in the regions < < and < <. Therefore, = cos() is also below the -ais in these regions. The minimum of = will become a maimum of =. WorKeD example Sketch = sin. Period: Amplitude: Horizontal shift: 0 Vertical shift: 0 (, ) (, ) 0 = = = = (, ) (, ) 0 = = = Use the graph of = sin to sketch =. WritE/draW sin = over the domain (, 0) (, 0) 0 Topic TrIgonoMeTr 09

55 Find the -intercepts and hence the vertical asmptotes for the reciprocal graph. The graph of = sin is above the -ais in the region 0 <. The graph of = sin will also be above the -ais in this region. A maimum value of = is reached in the original graph, meaning that a minimum of = will be reached in the reciprocal graph. The graph of = sin is below the -ais in the region < 0. The graph of = sin is also below the -ais in this region. The minimum of = will become a maimum of =. -intercepts occur at =, = 0 and =. These will be the vertical asmptotes for the reciprocal function. (, 0) (, 0) 0 = = = 0 0 = 0 0 = = 0 = = = 0 Maths Quest SPECIALIST MATHEMATICS VCE Units and

56 The graph of = cot( ) The graph of = tan() can be used to find the graph of = tan() The graph of = tan() is shown below. = = = = (or = cot()). In the portion of the graph shown, the -intercepts occur at,, 0, and. This means that the reciprocal function will have vertical asmptotes at =, =, = 0, = and =. = tan() has asmptotes at =, =, = and =. Therefore, the reciprocal function will have -intercepts at these positions. Remembering that sections of the graph that are above the -ais for = tan() will also be above the -ais for = and similarl for sections below the -ais, the tan() graph of = (or = cot ()) looks like this: tan() = = = 0 = = Topic Trigonometr

57 WorKeD example Use the graph of = tan to sketch = over the domain tan. WritE/draW Sketch = tan. 8 Period: Dilation: Horizontal shift: 0 Vertical shift: 0 The graph of = tan has asmptotes at = and =. These will be the -intercepts of the reciprocal function. Find the -intercepts for = tan and hence the vertical asmptotes for the reciprocal graph. (, 0) (, 0) 0 8 = = The -intercepts will be = and =. -intercepts occur at =, = 0 and =. These will be the vertical asmptotes for the reciprocal function. 8 (, 0) (, 0) 0 8 = = MaTHs QuesT specialist MaTHeMaTICs VCe units and

58 If we consider the region between = and = 0, the graph of = tan is initiall above the -ais between = and = and is then below the -ais. This will also be true for the reciprocal function. WorKeD example 5 In a similar fashion, the graph for = 0 to = can be obtained. Sketch the graph of = Use the graph of = sin + to find the graph of = Amplitude: Period: Horizontal shift: left Vertical shift: 0 Transformations of reciprocal trigonometric graphs Transformations can also be applied to the reciprocal trigonometric graphs. sin +. sin + WritE/draW (, ) + over the domain [, ] = 8 = 0 8 = 0 = (, ) = 0 = Topic TrIgonoMeTr

59 Consider the graph of = sin +. The asmptotes will occur at =, = and = 7. To graph = move = Eercise.8 PRactise sin + sin + up. +, (, ) = = = = = = 7 Graphs of reciprocal trigonometric functions WE Use the graph of = cos() to sketch = over the domain cos(). Use the graph of = sin() to sketch = over the domain sin(). WE Use the graph of = sin() to sketch = over the domain sin(). Use the graph of = cos() to sketch = over the domain cos(). 5 WE Use the graph of = tan() to sketch = over the domain tan(). Use the graph of = tan() to sketch = over the domain tan(). (, ) 7 WE Sketch the graph of = cot + + over the domain [, ]. Maths Quest SPECIALIST MATHEMATICS VCE Units and

60 8 Sketch the graph of = sec + over the domain [, ]. Consolidate Master.9 Units & AOS Topic Concept Graphs of inverse circular functions Concept summar Practice questions 9 Use the graph of = sin() to sketch = 0 Use the graph of = cos to sketch = over the domain [, ]. sin() cos Use the graph of = tan to sketch = tan Sketch = sec() + over the domain [0, ]. Sketch = cosec Sketch = cot over the domain [0, ]. over the domain [, ] 5 Sketch = sec() over the domain [, ]. Sketch = sin + 7 Sketch = 0.5 cosec over the domain [, ]. over the domain [, ]. over the domain [, ]. over the domain [, ]. 8 Sketch = sec + over the domain [, ]. 9 Use the graph of = sin() + to sketch = over the domain sin() + 5, 5. Sketch both graphs on the same set of aes. Check our graphs with CAS. 0 a Use the graph of = cos () to sketch = over the domain cos (),. Sketch both graphs on the same set of aes. Check our graphs with CAS. b Hence, determine the graph of = tan () for the same domain. Graphs of inverse trigonometric functions There are at least two possible approaches to sketching inverse trigonometric functions. The first method is to find the inverse of the function (which will be a trigonometric function) and use our knowledge of trigonometric functions to sketch the trigonometric function and its inverse. Alternativel, ou could use our knowledge about transforming equations to transform = sin (), = cos () or = tan () as required. In the following worked eamples, we will find the original trigonometric function and then sketch both functions. Topic Trigonometr 5

61 WorKeD example 5 Sketch = sin (). WritE/draW Find the inverse of = sin (). = sin () = sin( ) Sketch = sin(). Amplitude: Period: Horizontal shift: 0 Vertical shift: 0 The domain needs to be restricted so that the function is one-to-one. The domain becomes,. The domain and range of = sin() become the range and domain of = sin () respectivel. 5 Use the graph of = sin() to sketch = sin () b reflecting the graph in the line =. = sin( ) Therefore, the inverse is = sin(). = sin () Restrict the domain to,. 0 = sin(): Domain,, range, = sin (): Domain,, range, 0 = MaTHs QuesT specialist MaTHeMaTICs VCe units and

62 WorKeD example Sketch = cos ( + 5) WritE/draW Find the inverse of = cos ( + 5). = cos ( + 5) + 5 = cos( ) = cos( ) 5 Therefore, = cos() 5. Sketch = cos() 5. Amplitude: Period: Horizontal shift: 0 Vertical shift: 5 down The domain needs to be restricted so that the function is one-to-one. The domain becomes [0, ]. The domain and range of = cos() 5 become the range and domain of = cos ( + 5) respectivel. = cos() Restrict the domain to [0, ]. 5 5 = cos() 5: Domain [0, ], range [, ] = cos ( + 5): Domain [, ], range [0, ] 0 0 Topic TrIgonoMeTr 7

63 5 Use the graph of = cos() 5 to sketch = cos ( + 5) b reflecting the graph in the line =. WorKeD example 7 Find the inverse of = tan (). Sketch = tan. Period: Horizontal shift: 0 Vertical shift: 0 Sketch = tan (). WritE/draW = tan () = tan () = tan Therefore = tan. 5 = (, ) = = 5 (, ) 8 MaTHs QuesT specialist MaTHeMaTICs VCe units and

64 The domain needs to be restricted so that the function is oneto-one. The domain becomes,. The domain and range of = tan become the range and domain of = tan () respectivel. 5 Use the graph of = tan to sketch = tan () b reflecting the graph in the line =. 5 = = tan : Domain, =, range R = tan (): Domain R, range, 5 = = 5 5 = = Topic Trigonometr 9

65 The net worked eample is completed b transforming the inverse trigonometric function. WorKeD example 8 The graph of Sketch = sin () +. = sin () + is the graph of = sin () raised b units. Sketch = sin (). Raise the graph b. This means that the range is now, WritE/draW = sin () = sin () 0 MaTHs QuesT specialist MaTHeMaTICs VCe units and

66 Eercise.9 Graphs of inverse trigonometric functions WE5 Sketch = cos (). PRactise Sketch = tan (). WE Sketch = sin ( + ). Sketch = tan ( ). 5 WE7 Sketch = sin (). Sketch = cos (). 7 WE8 Sketch = cos (). Consolidate Master Units & AOS Topic Concept 7 Transformations of inverse circular functions Concept summar Practice questions 8 Sketch = tan () +. 9 Sketch = sin. 0 Sketch = tan. Sketch = cos (). Sketch = tan ( ). Sketch = sin ( + ). Sketch = cos ( ). 5 Sketch = cos (). Sketch = sin (). 7 Sketch = tan () +. 8 Sketch = cos ( ) +. 9 a Draw the graph of = sec(). b Identif a suitable domain to make = sec() a one-to-one function. c Sketch the graph of = sec (). 0 Sketch = cot ( + ). Topic Trigonometr

67 ONLINE ONLY.0 Review the Maths Quest review is available in a customisable format for ou to demonstrate our knowledge of this topic. the review contains: short-answer questions providing ou with the opportunit to demonstrate the skills ou have developed to efficientl answer questions using the most appropriate methods Multiple-choice questions providing ou with the opportunit to practise answering questions using CAS technolog Etended-response questions providing ou with the opportunit to practise eam-stle questions. a summar of the ke points covered in this topic is also available as a digital document. REVIEW QUESTIONS Download the Review questions document from the links found in the Resources section of our ebookplus. studon is an interactive and highl visual online tool that helps ou to clearl identif strengths and weaknesses prior to our eams. You can then confidentl target areas of greatest need, enabling ou to achieve our best results. MaTHs QuesT specialist MaTHeMaTICs VCe units and