2 Trigonometric functions

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1 Theodore Voronov. Mathematics 1G1. Autumn 014 Trigonometric functions Trigonometry provides methods to relate angles and lengths but the functions we define have many other applications in mathematics..1 Measuring angles Angles can be measured in degrees and in radians. Notation: a or b rad. The notation for the unit rad is commonly omitted. (So if no units are indicated, that means radians.) A full circle is defined to be 360. In particular, it follows that a right angle (a quarter of a full circle) is 90. Radians are less arbitrary units of angle because they are defined in terms of arc length. An angle of 1 radian is defined to be the angle which makes an arc of length r on a circle of radius r. Since the total arc length of a circle is πr, there are π radians in a circle. So π rad = 360. Angles are normally measured anti-clockwise from the x-axis as indicated. This gives us a formula for converting between the two measurements. angle in degrees = 180 π For example 90 = π rads, 30 = π 6 rads.. Definitions of trigonometric functions (angle in radians) Let ABC be a right angled triangle containing the angle θ (exercise: draw and label a triangle to fit with the information below). We define sin(θ) = opposite hypotenuse = AC AB cos(θ) = adjacent hypotenuse = BC AB tan(θ) = opposite adjacent = AC BC. Note that tan(θ) = sin(θ) cos(θ). The main trigonometric functions, which are sine and cosine, are defined as above for an acute angle θ, i.e., for 0 θ π. However, we may notice that cos θ and sin θ are respectively the x- and the y-coordinates of a point on the unit circle. That immediately allows to extend them to other values of the angles and in particular note their periodicity: cos(θ + π) = cos θ, sin(θ + π) = sin θ, tan(θ + π) = tan θ. 1

2 Theodore Voronov. Mathematics 1G1. Autumn 014 Note some further useful relations: sin( π cos( π sin( π θ) = cos θ θ) = sin θ + θ) = cos θ cos( π + θ) = sin θ sin(π + θ) = sin θ sin(π θ) = sin θ cos(π ± θ) = cos θ All of them can be seen from the diagram of a unit circle at the xy plane. Sine and cosine values lie between -1 and +1. Tangent values can take any value (and are undefined for certain values of θ). Note that cos θ = sin ( π θ), sin θ = cos ( π θ). The angles θ and π θ are called complementary. (Hence the names: cosine, i.e., cosinus means complementi sinus.) Useful special values: sin(30 ) = cos(60 ) = 1, sin(60 ) = cos(30 ) = sin(45 ) = cos(45 ) = 1, tan(45 ) = 1 3 Example. Convert the following from radians to degrees: 3π 4 =.... Examples. Find the exact value of the following: (i) sin(5 o ) (ii) cos( π 3 ) (iii) tan( π 3 ) Besides the main trigonometric functions, sine, cosine and tangent (in fact, it is sufficient to consider only sine and cosine), there are also sec(θ) = 1 cos(θ) cosec(θ) = 1 sin(θ) cot(θ) = 1 tan(θ) Graphs of the trigonometric functions are as follows.

3 Theodore Voronov. Mathematics 1G1. Autumn f(θ) = sin θ 0 π 1 1 f(θ) = cos θ 0 π 1 4 f(θ) = tan θ π 4 We can see that the trigonometric functions are periodic, meaning that the functions repeat the same values over a fixed period. Sine and cosine have period π and tangent has period π. 3

4 Theodore Voronov. Mathematics 1G1. Autumn 014 Example Sketch the graphs of the functions sin(x) and cos(x π ). How do they 3 differ from sin(x) and cos(x)?. Sketch the graphs of cosec(x), sec(x) and cot(x). 3. Find all the solutions of the following equations for x in radians. (i) sin(x) = 1 (ii) cos(x) = 0 (iii) tan(x) = 1.3 Geometric applications Take any triangle (not necessarily right-angled) with angles A, B, C and sides a, b, c opposite to A, B, C respectively. Given partial information about certain angles and lengths of sides we can use sine and cosine to determine the missing angles and lengths. Theorem.1 (Sine Rule). a sin(a) = b sin(b) = c sin(c) (To prove the sine rule, draw the altitude from B to side b. Denote it by h. Then, for two a right-angled triangles, we have h = c sin A = a sin C. Hence = c. Similar argument sin A sin C gives the remaining equality.) This rule is useful if we know (i) two angles and a side or (ii) two sides and a non-included angle. Theorem. (Cosine Rule). a = b + c bc cos(a) (Later we shall explain this by using vector algebra.) This rule is useful if we know (i) three sides or (ii) two sides and an included angle. Examples. Calculate all the missing angles and sides in the following triangles ABC. 1. A = 45, B = 30, a = 10. a = 3, b =, c = 5 3. a = 4, b = 6, B = π/6 4. b = 5, c = 8, A = Trigonometric identities You should be familiar with the following result. Theorem.3 (Pythagoras Theorem). In a right-angled triangle, the sum of the square on the hypotenuse is equal to the sum of the squares on the other two sides, i.e., r = x + y where r is the length of the hypotenuse, x, y, the lengths of the other two sided. 4

5 Theodore Voronov. Mathematics 1G1. Autumn 014 We can use the definitions of the trigonometric functions, together with the Pythagoras Theorem to obtain the following main identities satisfied by trigonometric functions. Theorem.4. For all values of the argument θ, Proof. From the Pythagoras theorem we have cos θ + sin θ = 1. ( x ) ( y ) cos θ + sin x θ = + = r r r + y r = x + y r = r r = 1 Corollary.1. For all values of θ for which the functions are defined: 1 + tan θ = sec θ cot θ + 1 = cosec θ Proof. These formulas are obtained by dividing throughout by cos θ and sin θ respectively. Further identities are based on the following addition formulas (or compound angle formulas). Theorem.5 (Addition formulas). For any angles A and B, 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 Theorem.6 (Addition formulas for tangent). tan(a + B) = tan(a B) = Proof. Use the addition formula for sine and cosine: tana + tanb 1 tana tan B tan A tan B 1 + tana tan B tan(a + B) = sin(a + B) cos(a + B) = sin A cos B + sin B cos A cos A cos B sin A sin B = The second formula follows in the same way. sin A cos B sin B cos A + cos A cos B cos A cos B sin A sin B 1 cos A cos B = sin A + sin B cos A cos B sin B cos A cos B 1 sin A = tana + tanb 1 tana tan B 5

6 Theodore Voronov. Mathematics 1G1. Autumn 014 Theorem. (Double angle formulas). The latter identity may also be written Proof. Let θ = A = B in the sum identities sin θ = sin θ cos θ cos θ = cos θ sin θ cos θ = cos θ 1 = 1 sin θ. sin θ = sin (θ + θ) = sin θ cos θ + cos θ sin θ = sin θcos θ cos θ = cos (θ + θ) = cos θ cos θ sin θ sin θ = cos θ sin θ Using the identity cos θ + sin θ = 1 to eliminate either cos θ or sin θ from the identity for cos θ completes the proof. Corollary.. tan(x) = tan(x) 1 tan (x) It is possible to deduce general formulas for cos nx and sin nx. We shall not do that, but consider particular examples instead. Example.. cos 3x = cos(x + x) = cos x cos x sin x sin x = (cos x sin x) cos x sin x cos x sin x = ( cos x 1) cos x cos x(1 cos x) = 4 cos 3 x 3 cos x. Therefore cos 3x = 4 cos 3 x 3 cos x. Example.3. cos 4x = cos(3x + x) = (4 cos 3 x 3 cos x) cos x (3 sin x 4 sin 3 x) sin x = 4 cos 4 x 3 cos x 3 sin x + 4 sin 4 x = 4 cos 4 x + 4 sin 4 x 3 = 4 cos 4 x + 4(1 cos x) 3 = 4 cos 4 x + 4(1 cos x + cos 4 x) 3 = 8 cos 4 x 8 cos x + 1. Therefore cos 4x = 8 cos 4 x 8 cos x + 1. Example.4. By using the addition formula show that sin(5 ) = Solving trigonometric equations From the graphs of sin(x), cos(x) and tan(x) we can see that there are infinitely many values of x that give the same value of the function. This causes problems when we want to go back from the value of the function to the value of the angle that it came from - because there are infinitely many angles that could have given that value of the function. For example how do we solve sin(x) = 0? There are infinitely many solutions to this equation, x = 0, ±π, ±π,... 6

7 Theodore Voronov. Mathematics 1G1. Autumn 014 We can define an inverse operation to taking sine if we choose a standard interval of width π and restrict the values that this inverse can take to that interval. We ll choose our standard interval for the inverse sine function 1 arcsin to run from π to π (excluding π itself to avoid a repetition). That is, define arcsin(x) to be the unique value y with π < y π and sin(y) = x. So arcsin(0.5) = π rads, arcsin( 0.8) = 0.93 rads. 6 Similarly we can define inverse functions for cosine and tangent. The inverse cosine, arccos(x), of x is the unique value y with 0 y π and cos(y) = x. (Note we chose a different standard interval for this but, if you think about the shape of the graph of cos, it does make sense to do so.) The inverse tangent, arctan(x), of x is the unique value y with π < y < π and tan(y) = x. When we solve equations involving trigonometric functions we should consider all possible solutions initially, not just those one our chosen interval. Some of these may be later discarded due to physical restrictions on the range of possible values (e.g. we may need only positive solutions). For example consider the equation cos(5y) = 0.5 ( ) We have arccos( 0.5) = π 3 cos(x) = 0.5 for. From the graph of cos(x) we see that we get the same value of x = π 3 ± nπ and x = 4π 3 ± nπ for any n = 0, 1,,... So the solution to ( ) is y = π 15 ± nπ and y = 4π 5 15 ± nπ for any n = 0, 1,,... 5 This gives values y =..., 4π 15, π 15, π 15, 4π 15, 8π 15,... Example.5. Find all solutions to the following equations. (i) sin(4x) = 1 (ii) tan(3x) = 1 Example.6. Some frequently met values of the arc functions: arcsin 0 = 0, arccos 0 = π, arctan 0 = 0, arcsin = arccos = π, arcsin 3 = π, arcsin 1 = π, arctan 1 = π Example.. Let cos α = 1. Find all values of α. Solution: We have α = Arccos 1 = ± arccos 1 + kπ = ± π + kπ. Note that π + π = 5π, so the solution can be written in an alternative form α = π + kπ or α = 5π + kπ. 3 3 Example.8. Find θ in the range 0 θ < π such that tan θ = 5. Solution: The unique solution in the range between π and π is θ = arctan (using a calculator or tables). General solution of the equation tan θ = 5 is obtained by adding integral multiples of π. However, adding any multiple of π to to θ = arctan 5 takes it out of the range 0 θ < π. Hence the answer is: θ = arctan The notation sin 1 (x) is also used in place of arcsin(x) but we avoid this, and corresponding notations for the other inverse trig functions, here because that notation might (wrongly!) suggest the meaning 1/ sin(x).

8 Theodore Voronov. Mathematics 1G1. Autumn 014 Example.9. Solve for all values: cos x = 1. Solution. We have cos x = 1 or cos x = 1. Therefore x = ± arccos 1 + kπ or x = π ± arccos 1 + kπ. This combines into x = ± arccos 1 + kπ. Example.10. Solve the equation: sin x + 5 sin x 3 = 0. Solution. We factorize: ( sin x 1)(sin x + 3) = 0, so sin x = 1 or sin x = 3. The second equation has no solutions, so we have sin x = 1 x = ( 1)k arcsin 1 + kπ = ( 1)k π + kπ. 6 Example.11. Express cos x + 3 sin x in the form A sin(x + x 0 ) where A and x 0 are to be determined. Solution. We can write cos x + 3 sin x = ( ) cos x + sin x = ( cos x + 3 ) sin x Now we look for x 0 such that cos x 0 = 3 and sin x 0 =. We can take x 0 = arcsin = arccos 3. So cos x + 3 sin x = sin(x + x 0 ) where x 0 = arcsin. 8

2 Recollection of elementary functions. II

2 Recollection of elementary functions. II Recollection of elementary functions. II Last updated: October 5, 08. In this section we continue recollection of elementary functions. In particular, we consider exponential, trigonometric and hyperbolic