A-Level Mathematics TRIGONOMETRY G. David Boswell - R2S Explore 2019 1. Graphs the functions sin kx, cos kx, tan kx, where k R; In these forms, the value of k determines the periodicity of the trig functions. Since our axes are calibrated in a measure of angles, the periodicity of these functions is the angle covered for one cycle. (In many applications, the horizontal axes is calibrated in units of time and the periodicity if then defined as a frequency). Some properties of the trig functions are: Periodicity The Sine and Cosine functions have a periodicity of 2π radians (360 degrees) The Tangent function has a periodicity of π radians (180 degrees). Symmetry The Sine and Tangent functions are odd functions (recall the definition that if f ( x) = f (x), then f (x) is odd). Therefore, the y-axis is not their mirror line of symmetry. The Cosine function is an even function (recall the definition that if f ( x) = f (x), then f (x) is even). Hence, the y-axis is a mirror line to symmetry. Amplitude The Sine and Cosine functions have an amplitude of 1 unit. The overall variation of their amplitude is ±1 unit. They are bounded between the lines y = 1 and y = -1. The Tangent function has infinite amplitude; it is unbounded. This is because tan(π / 2) + when we approach θ = π / 2 from the left and tan(π / 2) when we approach θ = π / 2 from the right. This we will see from the graphs. Asymptotes The Sine and Cosine functions does not have any vertical asymptotes. The Tangent function has an infinite number of vertical asymptotes starting at θ = π / 2 and separated by π radians left and right along the x-axis. HAMPTON SCHOOL G. David BÖ ZïK Inc. EPSE / MSE Reference (subject to change) Page 1 of 17
Table 1.a Graphs the functions sin x, cos x, tan x (i.e., k = 1) for the 4 Quadrants 1 Function & Comments Graph (Domain: ) 0 θ 2π y = sin x With k = 1, Then 1 cycle is completed in 360 o (or 2π radians). So Period, 2π k = 2π 1 = 2π Here, y = sin x is positive in the 1st and 2nd quadrants. And, y = sin x has a range of [-1, 1]. y = cos x With k = 1, Then 1 cycle is completed in 360 o (or 2π radians). So Period, 2π k = 2π 1 = 2π Here, y = cos x is positive in the 1st and 4th quadrants. And, y = cos x has a range of [-1, 1]. y = tan x With k = 1, then 1 cycle is completed in 180 o (or π radians). So Period, π k = π 1 = π. Here, y = tan x is positive in the 1st and 3rd quadrants. And, y = sin x has a range of [,+ ]. Do you agree that x = π for all 2 ± nπ integers n are vertical asymptotes? (why?) 1 Apply the CAST nemonic for Sine, Cosine and Tangent (or reverse it for Cosecant, Secant and Cotangent) functions) HAMPTON SCHOOL G. David BÖ ZïK Inc. EPSE / MSE Reference (subject to change) Page 2 of 17
Table 1.b Graphs the functions sin x, cos x, tan x (i.e., k = 2) Function & Comments Graph (Domain: ) 0 θ 2π y = sin2x With k = 2, then 2 cycles are completed in 360 o (or 2π radians). So Period = 2π k = 2π 2 = π. That is, 1 cycle is completed in 180 o. y = cos2x With k = 2, then 2 cycles are completed in 360 o (or 2π radians). So Period = 2π k = 2π 2 = π. That is, 1 cycle is completed in 180 o. y = tan2x Here, 2 cycles are completed in 180 o (or π radians) since this is the periodicity of the tangent function. Hence, 1 cycle is completed in 90 o. So, Period = π k = π 2. Can you see that x = π for all 4 ± n π 2 integer values of n are all vertical asymptotes? HAMPTON SCHOOL G. David BÖ ZïK Inc. EPSE / MSE Reference (subject to change) Page 3 of 17
Table 1.b Graphs the functions sin x, cos x, tan x (i.e., k = ½) Function & Comments Graph (Domain: ) 0 θ 2π y = sin x 2 Here, ½ cycle is completed in 360 o (or 2π radians). This implies that 1 cycle would be completed in 720 o (or 4π radians). So, Period = 2π k = 2π 1/ 2 = 4π. The starting point is (0, 0). y = cos x 2 Here, ½ cycle is completed in 360 o (or 2π radians). This implies that 1 cycle would be completed in 720 o (or 4π radians). So, Period = 2π k = 2π 1/ 2 = 4π. The starting point is (0, 1). y = tan x 2 Here, ½ cycle is completed in 180 o (or π radians). This implies that 1 cycle would be completed in 360 o. So, Period = π k = π 1/ 2 = 2π. The starting point is (0, 0); is a vertical asymptote. Where x = π 2 are the others located? HAMPTON SCHOOL G. David BÖ ZïK Inc. EPSE / MSE Reference (subject to change) Page 4 of 17
2. Relate the periodicity, symmetries and amplitudes of the functions in Specific Objective 1 above to their graphs; We have covered these above and here are some examples. 2.a g(x) = 2sin( 3x) ( ) = 1 g max (x) = 2 ( +1) = 2 Range : g min (x) = 2 1 and. So, the range is [-2, 2] 1 Amplitude : = units (this you can get by inspection) 2 (g g ) 1 max min (2 ( 2)) = 2 2 2π Periodicity : 3 cycles in 360 o (or 2π radians). Therefore 1 cycle in 120 o (or rads.) 3 Vertical Shift : None Horiz. Mirror : g(x) = 0 (the x-axis) Horiz. Shift : None. ( ) Starting pt. : A possible point for this sine curve is 0, 0. Repeat 2(a) for: 2.b 2.c 2.d 2.e y(x) = 2.4 cos 5x + π 4 y(x) = tan( 18x 45 ) m(x) = sin( x +15 ) m(θ) =10 3cosθ HAMPTON SCHOOL G. David BÖ ZïK Inc. EPSE / MSE Reference (subject to change) Page 5 of 17
3. Use the fact that sin π. (This extends to other trig co-functions). 2 ± x = cos x Here, we have a statement that the sine of an angle is equal to cosine of its complement. This we proved using a simple right angle triangle (see your notes). 3.a Prove that sin π. 2 ± x = cos x PROOF: In a right angle triangle with adjacent side to angle x as a, opposite side b, and hypotenuse c, then cos x = a and sin π. Therefore,. c 2 x = a sin π c 2 x = cos x If we now replace x with x. Then we also get sin π 2 ( x) = cos( x) sin π 2 + x = cos x (since the cosine function is even) Combining these 2 results and we get sin π Q.E.D. 2 ± x = cos x Comment 1: sin π means that the cosine graph can be obtained by shifting the 2 + x = cos x sine graph 90 o to the left. Comment 2: sin π implies that 2 x = cos x sin x + π = cos x 2 sin x π = cos x 2 sin x π = cos x 2 (since sine is an odd function) This means that the cosine graph can be obtained by (i) shifting the sine graph 90 o to the right (horizontally) and then (ii) reflecting that result in the y-axis. Follow? HAMPTON SCHOOL G. David BÖ ZïK Inc. EPSE / MSE Reference (subject to change) Page 6 of 17
3.b Prove the cofunction identity that tan π. 2 x cot x PROOF: By using the compound angle identity, then LHS = tan π 2 x (use a definition of tangent) sin π 2 x = cos π 2 x (use the co-functions of sine & cosine) = cos x (use a definition of cotangent) sin x = cot x = RHS Thus, by the method of Direct Proof, tan π Q.E.D. 2 x cot x 3.c As an exercise left for the reader, below are a few problems that you should be able to do by now. Verify which of the following are identities: (i) sec π (That is, Are secant and cosecant functions co-functions? ) 2 x = csc x (ii) sin 9π 2 + x = cos x (iii) cos 3π 2 θ = sinθ (iv) sin 21π 2 x = cos x HAMPTON SCHOOL G. David BÖ ZïK Inc. EPSE / MSE Reference (subject to change) Page 7 of 17
4. Use the formulae for sin(a ± B), cos(a ± B) and tan(a ± B); These are the compound angle identities (see the notes above in Section B). At this stage, first principles proofs are not required. In summary, they are sin(a ± B) sin Acos B ± cos Asin B cos(a ± B) cos Acos B sin Asin B tan A ± tan B tan(a ± B) 1 tan A tan B Here are some mixed examples. 4.a Prove cofunction identity that sin π. 2 ± x cos x 4.b Finding the exact values of compound trig functions (i) sin( 45 + 30 ) = sin 45 cos 30 + cos 45 sin 30 = 2 2 = 2 4 (Notice that (i) and (ii) gave the same result. This is because the cofunction came into play such that: sin 45 + 30 3 2 + 2 1 2 2 ( 3 +1) (ii) cos( 45 30 ) = cos 45 cos 30 + sin 45 sin 30 = 2 2 = 2 4 3 2 + 2 1 2 2 ( 3 +1) ( ) = sin 75 = cos(90 75 ) = cos15 = cos(45 30 ) Now try these. Find the exact value of: (iii) tan( 240 + 45 ) (iv) tan 11π 6 π 4 (v) sin 5π 12 cos π 5π cos 4 12 sin π 4 ). HAMPTON SCHOOL G. David BÖ ZïK Inc. EPSE / MSE Reference (subject to change) Page 8 of 17
5. derive the multiple angle identities for sin ka, cos ka, tan ka, for k Q; Again, we use these compound angle identities sin(a ± B) sin Acos B ± cos Asin B, tan A ± tan B cos(a ± B) cos Acos B sin Asin B and tan(a ± B) where we replace B 1 tan A tan B with (k-1)a as needed. Some examples are as follows. 5.a sin2a = sin(a + B) B A = sin(a + A) = sin Acos A + cos Asin A = 2sin Acos A 5.b cos2a = cos(a + B) B A = cos(a + A) = cos Acos A + sin Asin A = cos 2 A sin 2 A (1 of 3 forms) = cos 2 A (1 cos 2 A) = 2cos 2 A 1 (2 of 3 forms) = (1 sin 2 A) sin 2 A = 1 2sin 2 A (3 of 3 forms) 5.c tan2a = tan(a + A) Now, derive the identities for the following: 5.d sin 3A 5.e cos3a. 5.f tan3a 5.g cos5a tan A + tan A = 1 tan A tan A = 2 tan A 1 tan 2 A HAMPTON SCHOOL G. David BÖ ZïK Inc. EPSE / MSE Reference (subject to change) Page 9 of 17
6. Derive the identity sin 2 θ + cos 2 θ 1 As mentioned in the notes, this is a direct application of Pythagoras theorem. 6.a. PROOF: Consider the right angle triangle as shown. Then, Pythagoras theorem, y a 2 + b 2 c 2 such that c b O a x a 2 c + b2 2 c 1 2 a 2 c + b 2 c 1 But, by definition of the trig ratios, irrespective of where the triangle is positioned in the circle, then, and. sinθ = b cosθ = a tanθ = a c c b Therefore, by substitution of these ratios into the modified Pythagoras theorem, we get sin 2 θ + cos 2 θ 1 Q.E.D. 6.b. Assignment : Prove the other 2 identities based on Pythagoras theorem using a similar approach. (i) 1+ tan 2 θ sec 2 θ (ii) 1+ cot 2 θ csc 2 θ 7. Use the reciprocal functions sec x, cosec x and cot x We use these in problems as we go along. So no need for isolated examples. HAMPTON SCHOOL G. David BÖ ZïK Inc. EPSE / MSE Reference (subject to change) Page 10 of 17
8. Derive the corresponding identities for tan 2x, cot 2x, sec 2x and cosec 2x; Let us start with one that we have done before. 8.a tan2a = tan(a + A) = tan A + tan A 1 tan A tan A So, tan2a 2 tan A 1 tan 2 A So, by definition of the reciprocal trigonometric function, we derive the cotangent function as a double angle identity in 8.b. 8.b cot 2A = ( tan2a) 1 2 tan A = 1 tan 2 A 1 = 1 tan2 A 2 tan A 1 1 = cot 2 A 2 cot A (then we change from tan A to cot A) (multiply throughout by cot 2 A) So, cot 2A cot2 A 1 2cot A 8.c Prove that sec 2A sec 2 A 2 sec 2 A 8.d. Reader s task ~ use a similar approach as in 8.e to develop an identity for csc2a. HAMPTON SCHOOL G. David BÖ ZïK Inc. EPSE / MSE Reference (subject to change) Page 11 of 17
9. Develop and use the expressions for sin A ± sin B, cos A ± cos B; For the very simple proofs, please refer to page 7 of this document. Nonetheless, I have re-summarize the results here but replacing C and D with A and B, respectively. These are the Factor (or Sum-to-Product) formulae for Sine and Cosine based on the compound angle identities. sin A + sin B = 2sin A + B 2 cos A B 2 Sines: sin A sin B = 2cos A + B 2 sin A B 2 cos A + cos B = 2cos A + B 2 cos A B 2 Cosines: cos A cos B = 2sin A + B 2 sin A B 2 9.a Verify that cos2xsin5x = 1 ( sin 7x + sin 3x ) 2 9.b Write sin5θ sinθ as the product of two functions. 9.c Simplify sin6x sin2x sin6x + sin2x 9.d Prove that cot 4x cos2x sin2x sin 4x = tan 3x 9.e Derive the Product-to-Sum formula from cos A + cos B = 2cos A + B cos A B 2 2 Hence or otherwise, find the exact value of cos 75 cos15. HAMPTON SCHOOL G. David BÖ ZïK Inc. EPSE / MSE Reference (subject to change) Page 12 of 17
10. Use Specific Objectives 3, 4, 5, 6, 7, 8 and 9 above to prove simple identities. A separate set of proofs (solutions) was provided for these problems: 10.1 sin A tan A + cos A = sec A 10.11 csc A cot A + tan A = cos A 10.12 ( sin A + csc A) 2 = sin 2 A + cot 2 A + 3 11. Express f (θ) in the form r cos(θ ± α) and r sin(θ ± α) where r is positive 0 α < π / 2 Here, we are applying the definitions and compound angle identities to solve problems. r cos(θ ± α) and r sin(θ ± α) are trig functions with amplitude r and phase shift ±α. ( ) 11.a Express 4 cos x 5sin x in the form Rcos x + α ( ) 11.b Express 4 cos x + 8sin x in the form Rcos x α. 12. Find the general solution of equations of forms 1, 2, 3 and 4 below (i) Form 1: sin kθ = c, Solution Approach Since the general sine function has a periodicity of 2π radians, then sin kθ = c implies that kθ = ( sin 1 c)± 2nπ θ = 1 k ( sin 1 c)± 2nπ This is the general solution of sin kθ = c. 12.a Find the general solution of 5sin x = 4 HAMPTON SCHOOL G. David BÖ ZïK Inc. EPSE / MSE Reference (subject to change) Page 13 of 17
(ii) Form 2: cos kθ = c, Solution Approach Similarly, since the general cosine function has a periodicity of 2π radians, then cos kθ = c implies that kθ = ( cos 1 c)± 2nπ θ = 1 ( cos 1 c)± 2nπ k This is the general solution. 12.b Find the general solution of 10cos 3x = 5 (iii) Form 3: tan kθ = c, Solution Approach The tangent function has a periodicity of π radians such that tan kθ = c implies that kθ = ( tan 1 c)± nπ θ = 1 ( cos 1 c)± nπ k This is the general solution. 12.c Find the general solution of 3tan5x = 3 The following are left up to the reader: 12.d Solve 3sin x 2 = 0 for 0 x 360 correct to 2 decimal places. 12.e Find the general solution of 3tan2x = 2 12.f Solve 2sin 2 x cos x cos x = 0 for 0 x < 2π (work in rads.; leave π in your ans.) HAMPTON SCHOOL G. David BÖ ZïK Inc. EPSE / MSE Reference (subject to change) Page 14 of 17
(iv) Form 3: asinθ + bcosθ = c, for a, b, c, k, R; 12.g Using the results of 11.a above, find the general solution of 4 cosθ 5sinθ = 3. 12.h Find the general solution of 8sinθ + 6cosθ = 7. 12.i Reader: Find the general solution of 7cos x + 6sin x = 2 using the above approach. 12.j Reader: Find the general solution of 7cosθ 6sinθ = 2 using the above approach. 13. Find the solutions of the equations in 12 above for a given range; Consider this general result: θ =10.8 ± 360n, 246.6 ± 360n where n = 0,1,2,3... If the results are constrained, then substitute increasing integer values of n starting from 0 until all possible results are obtained within the required range. So, lets say 0 θ 540, then With n = 0: θ =10.8, 246.6 With n = 1: θ =10.8 + 360 = 370.8, 246.6 + 360 = 606.6 At this stage, there is no no need to go further as we will exceed the range of the solution space 0 θ 540 when n > 1. Also, the negative solutions immediately fall out of this range. So, the final solution set of angled would have been θ = 10.8, 246.6, 370.8. { } HAMPTON SCHOOL G. David BÖ ZïK Inc. EPSE / MSE Reference (subject to change) Page 15 of 17
14. Obtain maximum or minimum values of f (θ) for 0 θ 2π. 14.a As an example, consider Example 12.g where we can write 4 cosθ 5sinθ = 41cos(θ + 51.3 ) Since the cosine function is bounded between -1 and +1, then The maximum value of " 4 cosθ 5sinθ = " 41 (+1) = 41 and this occurs when cos(θ + 51.3 ) =1 θ + 51.3 = 0 θ = 51.3, that is, θ = α where α = phase shift Also, the minimum value of 4 cos x 5sin x = 41 ( 1) = 41 and this occurs when cos(θ + 51.3 ) = 1 θ + 51.3 =180 θ =128.7, that is, θ =180 α These results can be estimated from the graph of y = 4 cosθ 5sinθ. Reader ~ Please verify. HAMPTON SCHOOL G. David BÖ ZïK Inc. EPSE / MSE Reference (subject to change) Page 16 of 17
14.b Reader: Here is the final problem in parts (i). Show that f (x) = sin x + 3 cos x = 2sin x + 2π. 3 (ii) What are the solutions of 2sin x + 2π =1 for 0 x 2π? 3 (iii) What is the Max value of f (x) = sin x + 3 cos x and under what value of x? (iv) What is the Min value of f (x) = sin x + 3 cos x and under what value of x? (v) Estimate the y-intercept of f (x) = sin x + 3 cos x? (vi) Hence of otherwise, sketch the graph of f (x) = sin x + 3 cos x for 0 x 2π. - ENFIN - HAMPTON SCHOOL G. David BÖ ZïK Inc. EPSE / MSE Reference (subject to change) Page 17 of 17