Question

sin^–1(sin2) + sin^–1(sin4) + sin^–1(sin6) = ________

(a) π – 12

(b) 0

(c) 12 

(d) 12 – π

Answer 1

The correct option  (b) 0  

Explanation:

sin^–1(sin2) + sin^–1(sin4) + sin^–1(sin6)

Let 2 = π – θ₁, 4 = π + θ₂ & 6 = 2π – θ₃ considering quadrants in which angles lie, 

∴ given:

sin^–1[sin(π – θ₁)] + sin^–1[sin(π + θ₂)] + sin^–1[sin(2π – θ₃)]

= sin^–1(sinθ₁) + (– 1)sin^–1(sinθ₂) + (– 1) sin^–1(sinθ₃)

= θ₁ – θ₂ – θ₃ 

= (π – 2) – (4 – π) + 6 – 2π 

= 0.