Book: RD Sharma - Mathematics (Volume 1)
Chapter: 4. Inverse Trigonometric Functions
Subject: Maths - Class 12th
Q. No. 22 of MCQ
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Choose the correct answer
If 
, then one of the possible values of θ is
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°
