Let θ = sin-1 (sin (-600°)), then the value of θ is:1. π/3 2. �

1.	Let θ = sin-1 (sin (-600°)), then the value of θ is:1. π/3 2. π/23. 2π/34. -2π/3
Answer» Correct Answer - Option 1 : π/3 Explanation: θ = sin-1 (sin (-600°)) θ = sin-1 (sin (-600°)) θ = sin-1 (sin (-600°)) ----------(1) Since, sin (θ + (n × 360°)) = sin θ where n is a natural number Continuing from equation (1) θ = sin-1 (sin (-600°)) θ = sin-1 (sin (-600° + (2 × 360°))) θ = sin-1 (sin (-600° + 720°)) θ = sin-1 (sin 120°) θ = \(sin^{-1}(\frac{\sqrt{3}}{2})\) θ = π/3 Note: Range of inverse sine function is [-π/2 to π/2].

Let θ = sin-1 (sin (-600°)), then the value of θ is:1. π/3 2. π/23. 2π/34. -2π/3

Answer» Correct Answer - Option 1 : π/3

Explanation:

θ = sin-1 (sin (-600°))

θ = sin-1 (sin (-600°)) ----------(1)

Since, sin (θ + (n × 360°)) = sin θ

where n is a natural number

Continuing from equation (1)

θ = sin-1 (sin (-600°))

θ = sin-1 (sin (-600° + (2 × 360°)))

θ = sin-1 (sin (-600° + 720°))

θ = sin-1 (sin 120°)

θ = \(sin^{-1}(\frac{\sqrt{3}}{2})\)

θ = π/3

Note: Range of inverse sine function is [-π/2 to π/2].