Inverse Trigonometric Functions

Question

Prove the following :

$3 space sin to the power of negative 1 end exponent straight x equals sin to the power of negative 1 end exponent left parenthesis 3 straight x minus 4 straight x cubed right parenthesis comma space straight x element of open square brackets negative 1 half comma 1 half close square brackets$

Solution

Put .x = cosθ
∴L.H.S. = 3cos –1(cosθ) = 3 θ
R.H.S. = cos ^–1(4x³ – 3 x) = cos ^–1(4 cos³ θ – 3 cos θ)
= cos ^–1(cos 3 θ) = 3 θ
∴L.H.S. = R.H.S.