3. Topic 3 - Trigonometry

Page 1 TOPIC 3: TRIGONOMETRY 3.0 Radians and degrees π rad = 180° Example 1: Conversion between radians and degrees. (a)

1 2

π rad =

°

(e) 300°

=

rad

(b)

3 2

π rad =

°

(f) 270°

=

rad

°

(g) 90°

=

rad

°

(h) 45°

=

rad

(c) 2π rad = (d)

3 4

π rad =

3.1.1 Secant θ, cosecant θ, cotangent θ (sec θ, cosec θ, cot θ) (a) sin θ =

(d) cosec θ =

(b) cos θ =

(e) sec θ =

(c) tan θ =

(f) cot θ =

Trigonometry identity cos2θ + sin2θ = 1 cot2θ + 1 = cosec2θ 1 + tan2 θ = sec2 θ Example 2 Find the exact values of (a) sec

2 5  2 , (b) cosec , (c) cot   3 6  3

3.1.2 Quadrants and angles Complementary angles sin (90° – θ) = cos θ, cos (90° – θ) = sin θ, tan (90°– θ) = 1/ tan θ = cot θ Positve angles in all four quadrants sin θ = sin (180° - θ) = - sin (180°+ θ) = - sin (360° - θ) cos θ = - cos (180° - θ) = - cos (180°+ θ) = cos (360° - θ) tan θ = - tan (180° - θ) = tan (180°+ θ) = - tan (360° - θ)

3   , (d) sec 4 

Page 2 Negative angles in all four quadrants sin θ = - sin (- θ) = -sin [-(180° - θ)] = sin [-(180° + θ)] = sin [-(360° - θ)] cos θ = cos (- θ) = - cos [-(180° - θ)] = - cos [-(180° + θ)] = cos [-(360° - θ)] tan θ = - tan (- θ) = tan [-(180° - θ)] = - tan [-(180° + θ)] = tan [-(360° - θ)] Example 3 List down all the angles between -2π and 4π with the same secant value as 52  . Example 4 Given that sin x = (i)

cosec2 x,

2 , where x is obtuse, find the exact value of 5 (ii) cos2 x, (iii) sec x, (iv)cot x

Testing your skills P. 68, Exercise 5A 2, 3, 5, 6. Exercise 3.1: Secant, Cosecant, Cotangent 1. Simplify the following: (a) sec(π – x) cos x (b) sin x (c) sec (- x) (d) 1 + tan2 x (e) cot (π + x) (f) cosec (π + x) 2. (a) (b) (c) (d) (e) (f) (g) (h)

Find the exact value of: sec 14  cosec 12 π cot 56  cosec (- 34  ) cot(- 13  ) sec (- 136  ) cot(  112  ) sec 76 

3. Given that sin A  53 , where A is acute, and cos B   12 , where B is obtuse, find the exact values of (a) sec A, (b) cot A, (c) cot B, (d) cosec B. 4. Given that cosec C = 7, sin2 D = 12 and tan2 E = 4, find the possible values of cot C, sec D and cosec E, giving your answers in exact form.

Page 3 3.2 Graphs of trigonometric functions and its applications (a) y = sin θ

(b) y = cosec θ = (sin θ)–1

(c) y = cos θ

(d) y = sec θ = (cos θ)–1

Page 4 3.2 (cont.) Graphs of trigonometric functions and its applications (e) y = tan θ = (sin θ) / (cos θ)

(f) y = cot θ = (cos θ) / (sin θ)

Applications of the graph to find the number of solutions for an equation Example 5 Given the θ is in the interval of -360° ≤ θ ≤ 720°, how many values of θ satisfy the equation cos θ = cos 20°. Example 6 Given the range θ is in the interval of 0° ≤ θ ≤ 360°, state the number of values of θ satisfying the equation cosec θ = cosec (– 30°). Example 7 State all the solutions of the equation: cot θ = cot(-150°), where θ is in the interval 180° ≤ θ ≤ 360°. Example 8 By sketching suitable graph, show that the equation sec x = 3 – x2 has exactly one root in the interval 0 < x < ½ π. Hence, state the number of solution in the interval –π < x < π.

-

Page 5 Exercise 3.2: Using graphs of trigonometry 1. The diagram shows a curved rod AB of length 100 cm which forms an arc of a circle. The end points A and B of the rod are 99 cm apart. The circle has radius r cm and the arc AB subtends an angle of 2α radians at O, the centre of the circle. (i) Show that α satisfies the equation 99 x  sin x . [3] 100 (ii) By sketching suitable graph, show that this equation has exactly one root in the interval 0 < x < ½ π. [2] (iii)Hence, state the number of root of this equation in the interval -π < x < π. [2]

2. By sketching a suitable graph of graphs, show that the equation cosec x = ½ x + 1, where x is in radians, has a root in the interval 0 < x < ½ π.

[2]

3. By sketching a suitable pair of graphs, show that the equation 2 cot x = 1 + ex, where x is in radians, has only one root in the interval 0 < x < ½ π.

[2]

4. By the graph of the trigonometric ratios from note 3.2, state the number of solutions for the following equations for -180° ≤ θ ≤ 720°. (a) tan θ = tan 70° (b) sin θ = sin -10° (c) sec θ = sec 180° (d) cosec θ = cosec 35°

Page 6 3.3 Trigonometric Functions On Sum And Difference Of Angles (i) (ii) (iii) (iv) (v) (vi)

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) = 1  tan A tan B tan A  tan B tan (A – B) = 1  tan A tan B

Special angles: θ = 30°

θ = 45°

θ = 60°

sin 30° =

sin 45° =

sin 60° =

cos 30° =

cos 45° =

cos 60° =

tan 30° =

tan 45° =

tan 60° =

Example 9 Find the exact value of (a) sin 15° and (b) tan 105°. Example 10 Without using a calculator, find the exact value of 1  tan 15  1 (b) (a) (cos 75   sin 75  ) 1  tan 15  2 Example 11

8 12 1 , that sin B = , and that 0 < B <  < A < π. Find the 17 13 2 exact value of tan (A + B) and cos (A – B) You are given that sin A =

Page 7 Exercise 3.3: Trigonometric Functions On Sum And Difference Of Angles 1. Find the exact value of (a) sin 75° (b) cos 105° (c) tan (-15°) (d) cot 75° 2. Find the value of : (a) sin 80° cos 70° + cos 80° sin 70° (b) cos 105°cos 15° + sin 105°sin 15° (c)

(d)

1

cos 15° –

1

sin 15° 2 2 3 1 sin60 + cos 60° (e) 2 2

tan 40   tan 20  1  tan 40  tan 20 

12 4 and sin B = , angles A and B are acute, find 13 5 (d) sin (A + B) (a) cos A (b) cos B (e) cos (A + B) (c) tan B (f) tan (A + B)

3. If sinA =

1 5 and cos B =  , angles A is acute and B is obtuse, find 7 5 (c) sin (A – B) (a) sin A (b) sin B (d) cos (A – B)

4. If cos A =

3.4: Solving equations with double angles, sec2θ and cosec2θ Double angles sin 2A = 2sin A cos A cos 2A = cos2 A – sin2 A = 2cos2 A – 1 = 1 – 2sin2 A 2 tan A tan 2 A  1  tan 2 A Trigonometry identities sin2 θ + cos2 θ = 1 tan2 θ + 1 = sec2 θ 1 + cot2 θ = cosec2 θ Example 12 Solve the equation 1 – 2sin θ – 4 cos 2θ = 0, for all values of θ from 0° to 360°. Example 13 Solve the equation tan 2x + 5 tan x = 0 for all values of θ from 0° to 360°. Example 14 Given that θ is obtuse with sin θ =

2 , find cos 2θ and tan 2θ. 3

Page 8 Exercise 3.4: Solving double angles’ trigonometric equations 1. Solve the following equations for 0° ≤ x ≤ 360°. (a) cos 2x = 5 cos x + 2 (b) 3 cos 2x + 1 = 2 sinx (c) cos x + 3 cos 2x = 2 (d) 4 sin x = 7 tan 2x 2. Given that x is acute such that cos x =

3 , find the exact values of 5

(a) sin 2x (b) cos 2x 3. Given that cos B =

3 , find the exact values of cos 2B and cos ½ B. 4

7 , find the possible values of cos A and sin A. 18 12 5. If tan 2A = , find the possible values of tan A. 5

4. If cos 2A =

6. The diagram shows a sector OAB of a circle with centre O and radius r. The angle AOB is α radians, where 0 < α < ½ π. The point N on OA is such that BN is perpendicular to OA. The area of the triangle ONB is half the area of the sector OAB. (i) Show that α satisfies the equation sin 2x = x. (ii) By sketching a suitable pair of graphs, show that this equation has exactly one root in the interval 0 < x < ½ π. [5] 3.5 Expression a sin θ + b cos θ in the form R sin(θ + α) or R cos (θ + α)

b : a (i) a sin θ + b cos θ ≡ R sin(θ + α), (ii) a sin θ – b cos θ ≡ R sin(θ – α), (iii) a cos θ + b sin θ ≡ R cos(θ – α), (iv) a sin θ – b cos θ ≡ R cos(θ + α) Given that R =

a 2  b 2 , tan  

Example 15 Solve the equation 3 cos θ – sin θ = 1, for 0° ≤ θ ≤ 360°. Hence find the solutions of the equation, giving your answer to the nearest 0.1°. Example 16 Express 3 cos θ + 4 sin θ, in the form R cos(θ – α). Hence find the: (a) maximum value of 3 cos θ + 4 sin θ (b) minimum value of 3 cos θ + 4 sin θ 1 (c) minimum value of 3 cos   4 sin   8

Page 9 Ex 3.5: Expression a sin θ + b cos θ in the form R sin(θ + α) or R cos (θ + α) 1. Solve the following equation for 0° ≤ θ ≤ 360°. (a) 8 sin θ – 15 cos θ – 10 = 0 (b) 3 cos θ + 7 sin θ = – 3 2. Express in the given form, and hence find the maximum and minimum values of (a) sin θ + 2 cos θ; Rsin(θ + α) (b) 2cos θ – sin θ; Rcos (θ + α) 3. Express cos x + ( 3 )sin x in the form R cos (x – α), giving the exact values of R and α such that R > 0, and 0° ≤ α ≤ 90°. Hence find the solutions of the equation cos x + ( 3 )sin x = 2 , giving your answer exactly in degrees. 4. Express 3 cos θ – 5 sin θ in the form R cos( θ + α), where R > 0, and 0° ≤ α ≤ ½ π. Hence, or otherwise, find the solutions of the equation 3 cos θ – 5 sin θ = 2, giving your answer correct to 3 significant figures. 5. Express 5 cos θ + sin θ in the form R cos( θ – α), where R > 0, and 0° ≤ α ≤ 90°. Hence state the maximum value of 5 cos θ + sin θ. 6. Express 8 sin x + 6 cos x in the form R sin ( θ + α), where R > 0, and 0° ≤ α ≤ ½ π. Hence state the minimum value of 8 sin x + 6 cos x. 7. Express 4 sinθ – 3 cos θ in the form R sin(θ – α), where R > 0, and 0° ≤ α ≤ 90°. Stating the value of α correct to 2 decimal places. Hence solve the equation 4 sin θ – 3 cos θ = 2, giving all values of θ such that ≤ θ ≤ 360°. 1 Write down the greatest value of . 4 sin   3 cos   6

0°

8. Express 5 sin θ + 12 cos θ in the form R sin (θ + α), where R is positive and α is acute, giving the value of α to the nearest 0.1°. Hence solve the equation 6 sec θ – 5 tan θ = 12, for values of θ lying between 0° and 360°, giving your answer to the nearest 0.1°.

Page 10 3.6 Proving Trigonometrical Identities (i) (ii) (iii) (iv)

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 (v) tan (A + B) = 1  tan A tan B tan A  tan B (vi) tan (A – B) = 1  tan A tan B (vii) sin 2A = 2sin A cos A (viii) cos 2A = cos2 A – sin2 A = 1 – 2sin2 A = 2cos2 A – 1 2 tan A (ix) tan 2 A  1  tan 2 A Example 17 (a) Prove that cos θ – cos 3θ = 4 sin2 θ cos θ. (b) Prove that cos 3θ = 4 cos3 θ – 3 cos θ. (c) Prove the identity cot x– cot 2x = cosec 2x. (d) Prove that sin (A + B) + sin (A – B) = 2 sin A cos B. Ex 3.6: Proving Trigonometrical Identities 1. Prove tan θ + cot θ = 2 cosec 2θ [3] 2. Prove that cos 2θ + tan θ sin 2θ = 1. [4] By letting θ = 15°, find the exact value of tan 15°. 2 3. Prove that identity tan A + cot A = . Hence, or otherwise, solve the equation sin 2 A tan (θ + 45°) + cot (θ + 45°) = 4, giving all the solutions in the interval 0° ≤ θ ≤ 360°. [5] 4. Prove the identity cot θ – tan θ = 2 cot 2θ. [3] 5.

6.

7.