Chapter 3 – Trigonometric Functions

A circular wire of radius 3 cm is cut and bent so as to lie along the circumference of a hoop whose radius is 48 cm. Find the angle in degrees which is subtended at the centre of hoop. Solution Given that circular wire is of radius 3 cm, so when it is cut then its length = 2π × 3 = 6π cm. Again, it is being placed along a circular hoop of radius 48 cm. Here, s = 6π cm is the length of arc and r = 48 cm is the radius of the circle. Therefore, the angle θ, in radian, subtended by the arc at the centre of the circle is given by Arc 6π π θ= = = = 22.5° . Radius 48 8

If A = cos2θ + sin4θ for all values of θ, then prove that ≤ A ≤ 1. Solution We have A =cos2 θ + sin4 θ = cos2 θ + sin2 θ sin2 θ ≤ cos2 θ + sin2 θ Therefore, A≤1 Also, A = cos θ + sin θ = (1 – sin2 θ) + sin4 θ

Find the value of 3 cosec 20° – sec 20° Solution We have 3 1

If θ lies in the second quadrant, then show that 1 − sin θ 1 + sin θ + = −2sec θ 1 + sin θ 1 − sin θ Solution We have 1 − sin θ 1 + sin θ 1 − sin θ 1 + sin θ 2 + = + = 1 + sin θ 1 − sin θ 1 − sin 2 θ 1 − sin 2 θ cos 2 θ = (Since α 2 = | α | for every real number α ) | cos θ | Given that θ lies in the second quadrant so |cos θ| = – cos θ (since cos θ < 0). Hence, the required value of the expression is = –2 secθ − cos θ

Find the value of tan 9° – tan 27° – tan 63° + tan 81° Solution We have tan 9° – tan 27° – tan 63° + tan 81° = tan 9° + tan 81° – tan 27° – tan 63° = tan 9° + tan (90° – 9°) – tan 27° – tan (90° – 27°) = tan 9° + cot 9° – (tan 27° + cot 27°) (1) 1 2 Also tan 9° + cot 9° = sin 9° cos9° = sin18° (Why?) (2) 1 2 2 Similarly, tan 27° + cot 27° = = = (Why?) (3) sin 27° cos 27° sin 54° cos36° Using (2) and (3) in (1), we get 2 2 2× 4 2× 4 tan 9° – tan 27° – tan 63° + tan 81° = – = – =4 sin 18° cos36° 5 −1 5 +1 sec8 θ − 1 tan 8 θ

Prove that = sec 4 θ − 1 tan 2 θ sec8 θ − 1 (1 − cos8 θ) cos 4 θ Solution We have = sec 4 θ − 1 cos8 θ (1 – cos 4 θ) 2sin 2 4 θ cos 4 θ = (Why?) cos8 θ 2sin 2 2 θ sin 4 θ (2 sin 4 θ cos 4 θ) = 2 cos8 θ sin 2 2 θ sin 4 θ sin 8 θ = (Why?) 2 cos8 θ sin 2 2 θ 2sin 2 θ cos 2 θ sin 8 θ = 2 cos8 θ sin 2 2 θ tan8 θ = (Why?) tan 2 θ

Solve the equation sin θ + sin 3θ + sin 5θ = 0 Solution We have sin θ + sin 3θ + sin 5θ = 0 or (sin θ + sin 5θ) + sin 3θ = 0 or 2 sin 3θ cos 2θ + sin 3θ = 0 (Why?) or sin 3θ (2 cos 2θ + 1) = 0 or sin 3θ = 0 or cos 2θ = – nπ When sin 3θ = 0, then 3θ = nπ or θ = 1 2π 2π π When cos 2θ = – = cos , then 2θ = 2nπ ± or θ = nπ ± 2 3 3 3 π π which gives θ = (3n + 1) or θ = (3n – 1) 3 3 nπ All these values of θ are contained in θ = , n ∈ Z. Hence, the required solution set nπ is given by {θ : θ = , n ∈ Z}

Solve 2 tan2 x + sec2 x = 2 for 0 ≤ x ≤ 2π Solution Here, 2 tan2 x + sec2 x = 2 which gives tan x = ± TRIGONOMETRIC FUNCTIONS 43 1 π 7π If we take tan x = , then x = or (Why?) 3 6 6 −1 5π 11π Again, if we take tan x = , then x = or (Why?) 3 6 6 Therefore, the possible solutions of above equations are π 5π 7 π 11π x= , , and where 0 ≤ x ≤ 2π

Find the value of 1 + cos 1 + cos 1 + cos 1 + cos 8 8 8 8 π 3π 5π 7π Solution Write 1 + cos 1 + cos 1 + cos 1 + cos 8 8 8 8 π 3π 3π π = 1 + cos 1 + cos 1 + cos π − 1 + cos π −

If x cos θ = y cos (θ + ) = z cos ( θ + ), then find the value of 3 3 xy + yz + zx. 1 1 1 Solution Note that xy + yz + zx = xyz + + . x y z 2π 4π If we put x cos θ = y cos (θ + ) = z cos θ + = k (say). 3 3 k k k Then x= ,y= and z = cos θ 2π 4π cos θ + cos θ + 3 3 1 1 1 1 2π 4π so that + + = cos θ + cos θ + + cos θ + x y z k 3 3 1 2π 2π = [cos θ + cos θ cos − sin θ sin k 3 3 4π 4π + cos θ cos − sin θ sin ] 3 3 1 −1 3 1 3 = [cos θ + cos θ ( )− sin θ − cos θ + sin θ] (Why?) k 2 2 2 2 ×0=0 = Hence, xy + yz + zx = 0

If α and β are the solutions of the equation a tan θ + b sec θ = c, 2ac then show that tan (α + β) = . a − c2 Solution Given that atanθ + bsecθ = c or asinθ + b = c cos θ Using the identities, θ θ 2 tan 1 − tan 2 2 and cos θ = 2 sin θ = 2 θ 2 θ 1 + tan 1 + tan 2 2 TRIGONOMETRIC FUNCTIONS 45 θ θ a 2 tan c 1 − tan 2 We have, 2 +b = 2 θ θ 1 + tan 2 1 + tan 2 2 2 2 θ θ or (b + c) tan + 2a tan +b–c=0 2 2 θ α β Above equation is quadratic in tan and hence tan and tan are the roots of this 2 2 2 α β −2a α β b−c equation (Why?). Therefore, tan + tan = and tan tan = (Why?) 2 2 b+c 2 2 b+c α β α β + tan tan tan + 2 2 Using the identity = 2 2 α 1 − tan tan β 2 2 −2a α β b+c −2a − a We have, tan + = = = ... (1) 2 2 b−c 2c c 1− b+c Again, using another identity α+β α+β 2 tan tan 2 = , 2 1 − tan 2 α+β 2 − c 2ac We have tan ( α + β ) = = [From (1)] a2 a − c2 1− 2 Alternatively, given that a tanθ + b secθ = c ⇒ (a tanθ – c)2 = b2(1 + tan2θ) ⇒ a2 tan2θ – 2ac tanθ + c2 = b2 + b2 tan2θ ⇒ (a2 – b2) tan2θ – 2ac tanθ + c2 – b2 = 0 ... (1) Since α and β are the roots of the equation (1), so 2ac c2 − b2 tanα + tanβ = and tanα tanβ = a2 − b2 a 2 − b2 tan α + tan β Therefore, tan (α + β) = 1 − tan α tan β 2ac a − b2 2ac = 2 2 = c −b a − c2 2 2 a −b

Show that 2 sin2 β + 4 cos (α + β) sin α sin β + cos 2 (α + β) = cos 2α Solution LHS = 2 sin2 β + 4 cos (α + β) sin α sin β + cos 2(α + β) = 2 sin2 β + 4 (cos α cos β – sin α sin β) sin α sin β + (cos 2α cos 2β – sin 2α sin 2β) = 2 sin2 β + 4 sin α cos α sin β cos β – 4 sin2 α sin2 β + cos 2α cos 2β – sin 2α sin 2β = 2 sin2 β + sin 2α sin 2β – 4 sin2 α sin2 β + cos 2α cos 2β – sin 2α sin 2β = (1 – cos 2β) – (2 sin2 α) (2 sin2 β) + cos 2α cos 2β (Why?) = (1 – cos 2β) – (1 – cos 2α) (1 – cos 2β) + cos 2α cos 2β = cos 2α (Why?)

If angle θ is divided into two parts such that the tangent of one part is k times the tangent of other, and φ is their difference, then show that k +1 sin θ = sin φ k −1 Solution Let θ = α + β. Then tan α = k tan β TRIGONOMETRIC FUNCTIONS 47 tan α k or = tan β 1 Applying componendo and dividendo, we have tan α + tan β k +1 = tan α − tan β k −1 sin α cos β + cos α sin β k +1 or = (Why?) sin α cos β − cos α sin β k −1 sin (α + β) k +1 i.e., = (Why?) sin (α − β) k −1 Given that α – β = φ and α + β = θ. Therefore, sin θ k +1 k +1 sin φ = k – 1 or sin θ = k −1 sin φ

Solve cos θ + sin θ = Solution Divide the given equation by 2 to get 3 1 1 π π π cos θ + sin θ = or cos cos θ + sin sin θ = cos 2 2 2 6 6 4 π π π π or cos − θ = cos or cos θ − = cos (Why?) 6 4 6 4 π π Thus, the solution are given by, i.e., θ = 2mπ ± + Hence, the solution are π π π π 5π π θ = 2mπ + + and 2mπ – + , i.e., θ = 2mπ + and θ = 2mπ – 4 6 4 6 12 12

If tan θ = , then sinθ is −4 4 −4 4

• (A)but not
• (B)or 5 5 5 5 4 4
• (C)but not −
• (D)None of these 5 5 Solution Correct choice is B. Since tan θ = − is negative, θ lies either in second quadrant or in fourth quadrant. Thus sin θ = if θ lies in the second quadrant or sin θ = − , if θ lies in the fourth quadrant.

If sin θ and cos θ are the roots of the equation ax2 – bx + c = 0, then a, b and c satisfy the relation.

• (A)a2 + b2 + 2ac = 0
• (B)a2 – b2 + 2ac = 0
• (C)a2 + c2 + 2ab = 0
• (D)a2 – b2 – 2ac = 0 Solution The correct choice is (B). Given that sin θ and cos θ are the roots of the b c equation ax2 – bx + c = 0, so sin θ + cos θ = and sin θ cos θ = (Why?) a a 2 2 2 Using the identity (sinθ + cos θ) = sin θ + cos θ + 2 sin θ cos θ, we have b2 2c =1 + or a2 – b2 + 2ac = 0 a a

The greatest value of sin x cos x is

• (A)1
• (B)2
• (C)2
• (D)Solution (D) is the correct choice, since 1 1 sinx cosx = sin 2x ≤ , since |sin 2x | ≤ 1 . 2 2 Eaxmple 18 The value of sin 20° sin 40° sin 60° sin 80° is −3 5 3 1 (A) (B) (C) (D)

The value of cos cos cos cos is 5 5 5 5 1 −1 −1

• (A)
• (B)0
• (C)
• (D)16 8 16 Solution (D) is the correct answer. We have π 2π 4π 8π cos cos cos cos 5 5 5 5 1 π π 2π 4π 8π = 2 sin cos cos cos cos π 5 5 5 5 5 2 sin 1 2π 2π 4π 8π = sin cos cos cos π 5 5 5 5 (Why?) 2 sin 1 4π 4π 8π = sin cos cos π 5 5 5 (Why?) 4 sin 1 8π 8π = sin cos π 5 5 (Why?) 8sin 16π π sin sin 3π + 5 5 = = π π 16 sin 16 sin 5 5 π − sin = 5 π (Why?) 16 sin = − Fill in the blank :

If 3 tan (θ – 15°) = tan (θ + 15°), 0° < θ < 90°, then θ = _________ Solution Given that 3 tan (θ – 15°) = tan (θ + 15°) which can be rewritten as tan(θ + 15°) 3 = . tan(θ − 15°) 1 tan (θ + 15°) + tan (θ – 15°) Applying componendo and Dividendo; we get =2 tan (θ + 15°) − tan (θ – 15°) sin (θ + 15°) cos (θ − 15°) + sin (θ − 15°) cos (θ + 15°) =2 sin (θ + 15°) cos (θ − 15°) − sin (θ − 15°) cos (θ + 15°) sin 2θ = 2 i.e., sin 2θ = 1 (Why?) sin 30° π giving θ = State whether the following statement is True or False. Justify your answer 1−

“The inequality 2sinθ + 2cosθ ≥ 2 holds for all real values of θ” TRIGONOMETRIC FUNCTIONS 51 Solution True. Since 2sinθ and 2cosθ are positive real numbers, so A.M. (Arithmetic Mean) of these two numbers is greater or equal to their G.M. (Geometric Mean) and hence 2sin + 2 ≥ 2sin × 2cos = 2sin + cos sin θ + cos θ 1 1 1 sin θ + cos θ 2 2 ≥2 2 =2 2 1 π sin + θ 2 4 ≥2 π Since, –1 ≤ sin + θ ≤ 1, we have −1 1 2sin θ + 2cos θ 1− ≥ 2 2 ⇒ 2sin θ + 2cos θ ≥ 2 2 Match each item given under the column C1 to its correct answer given under column C2

C1 C2 1 − cos x x (a)

• (i)cot 2 sin x 2 1 + cos x x (b)
• (ii)cot 1 − cos x 2 1 + cos x (c)
• (iii)cos x + sin x sin x (d) 1 + sin 2x
• (iv)tan Solution 1 − cos x 2 sin 2 2 x (a) = = tan . sin x x x 2 2 sin cos 2 2 Hence (a) matches with (iv) denoted by (a) ↔ (iv) 1 + cos x 2sin 2 (b) = 2 = cot 2 x . Hence (b) matches with (i) i.e., (b) ↔ (i) 1 − cos x 2 x 2 2 sin 1 + cos x 2 cos 2 2 x (c) = = cot . sin x x x 2 2 sin cos 2 2 Hence (c) matches with (ii) i.e., (c) ↔ (ii) (d) 1 + sin 2 x = sin 2 x + cos 2 x + 2 sin x cos x = (sin x + cos x ) 2 = ( sin x + cos x ) . Hence (d) matches with (iii), i.e., (d) ↔ (iii)