Answer :

We are given that,

θ = sin-1 {sin (-600°)}


We know that,


sin (2π – θ) = sin (4π – θ) = sin (6π – θ) = sin (8π – θ) = … = -sin θ


As, sin (2π – θ), sin (4π – θ), sin (6π – θ), … all lie in IV Quadrant where sine function is negative.


So,


If we replace θ by 600°, then we can write as


sin (4π – 600°) = -sin 600°


Or,


sin (4π – 600°) = sin (-600°)


Or,


sin (720° – 600°) = sin (-600°) …(i)


[, 4π = 4 × 180° = 720° < 600°]


Thus, we have


θ = sin-1 {sin (-600°)}


θ = sin-1 {sin (720° – 600°)} [from equation (i)]


θ = sin-1 {sin 120°} …(ii)


We know that,


sin (π – θ) = sin (3π – θ) = sin (5π – θ) = … = sin θ


As, sin (π – θ), sin (3π – θ), sin (5π – θ), … all lie in II Quadrant where sine function is positive.


So,


If we replace θ by 120°, then we can write as


sin (π – 120°) = sin 120°


Or,


sin (180° - 120°) = sin 120° …(iii)


[, π = 180° < 120°]


Thus, from equation (ii),


θ = sin-1 {sin 120°}


θ = sin-1 {sin (180° - 120°)} [from equation (iii)]


θ = sin-1 {sin 60°}


Using property of inverse trigonometry,


sin-1 (sin A) = A


θ = 60°


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