Chapter 3 Trigonometric Functions Miscellaneous Exercise 3 Solutions

II: Solve the following
Question 1.
Find the principal solutions of the following equations :
(i) sin2θ =
Solution:
sin2θ =
Since, θ ∈ (0, 2π), 2∈ ∈ (0, 4π)

(ii) tan3θ = -1
Solution:
Since, θ ∈ (0, 2π), 3∈ ∈ (0, 6π)

… [∵ tan(π – θ) = tan(2π – θ) = tan(3π – θ)
= tan (4π – θ) = tan (5π – θ) = tan (6π – θ) = -tan θ]
∴ tan3θ = tan = tan = tan = tan

(iii) cotθ = 0
Solution:
cotθ = 0
Since θ ∈ (0, 2π),
cotθ = 0 = cot = cot (π + …[∵ cos(π + θ) = cotθ]
∴ cotθ = cot = cot
∴ θ = or θ =
Hence, the required principal solutions are

(ii) tan5θ = -1
Solution:

(iii) cot2θ = 0
Solution:

(ii) cos²θ = -1
Solution:
cos2θ = -1
This is not possible because cos2θ ≥ 0 for any θ.
∴ cos2θ = -1 does not have any solution.

(iii) 2 sinθ = 3
Solution:
2 sin θ = 3 ∴ sin θ =
This is not possible because -1 ≤ sin θ ≤ 1 for any θ.
∴ 2 sin θ = 3 does not have any solution.

(iv) 3 sin θ = 5
Solution:
3 sin θ = 5
∴ sin θ =
This is not possible because -1 ≤ sin θ ≤ 1 for any θ.
∴ 3 sin θ = 5 does not have any solution.

(ii) tan²θ = 3
Solution:
The general solution of tan²θ = tan²∝ is
θ = nπ ± ∝, n ∈ Z.
Now, tan²θ = 3 = ()²
∴ tan²θ = (tan)²…[∵ tan = ]
∴ tan²θ = tan²
∴ the required general solution is
θ = nπ ± , n ∈ Z.

(iii) sin θ – cosθ = 1
Solution:
∴ cosθ – sin θ = -1

(iv) sin²θ – cos²θ = 1
Solution:
sin²θ – cos²θ = 1
∴ cos²θ – sin²θ = -1
∴ cos2θ = cosπ …(1)
The general solution of cos θ = cos ∝ is
θ = 2nπ ± ∝, n ∈ Z
∴ the general solution of (1) is given by
2θ = 2nπ ± π, n ∈ Z
∴ θ = nπ ± , n ∈ Z

(ii) .
Solution:

(iii) a²sin (B – C) = (b²– c²) sinA
Solution:
By sine rule,
= = = k
∴ a = ksinA, b = ksinB, c = ksinC
RHS = (b²– c²) sin A
= (k²sin²B – k²sin²C)sin A
= k²(sin²B – sin²C) sin A
= k²(sin B + sin C)(sin B – sin C) sin A

= k²× sin (B + C) × sin (B – C) × sin A
= k²∙sin(π – A)∙sin(B – C)∙sinA … [∵ A + B + C = π]
= k²sin A∙sin (B – C)∙sin A
= (k sin A)²∙sin (B – C)
= a²sin (B – C) = LHS.

(iv) ac cos B – bc cos A = (a²– b²).
Solution:
LHS = ac cos B – bc cos A
= ac – bc
=(c²+ a²– b²) – (b²+ c²– a²)
= (c²+ a²– b²– b²– c²+ a²)
= (2a²– 2b²) = a²– b2 = RHS.

(v) .
Solution:

(vi) .
Solution:
By sine rule,

(vii)
Solution:
By sine rule,

(ii) cos^-1x = π + tan^-1, if x < 0.
Solution: