-->

Inverse Trigonometric Functions

Question

Wired Faculty App

Prove that: $3 \sin^{- 1} x = \sin^{- 1} (3 x - 4 x^{3}), X \in [- \frac{1}{2}, \frac{1}{2}]$

Solution

When $- \frac{1}{2} \leq x \leq \frac{1}{2}$

we have,

$- \frac{1}{2} \leq x \leq \frac{1}{2} \Rightarrow - \frac{π}{6} \leq θ \leq \frac{π}{6} \Rightarrow - \frac{π}{2} \leq 3 θ \leq \frac{π}{2} Also, - \frac{1}{2} \leq x \leq \frac{1}{2} \Rightarrow - 1 \leq 3 x - 4 x^{3} \leq 1 ∴ \sin 3 θ = 3 x - 4 x^{3} \Rightarrow 3 θ = \sin^{- 1} (3 x - 4 x^{3}) \Rightarrow 3 \sin^{- 1} x = \sin^{- 1} (3 x - 4 x^{3})$

For Daily Free Study Material Join wiredfaculty

Inverse Trigonometric Functions

Prove that: $3 \sin^{- 1} x = \sin^{- 1} (3 x - 4 x^{3}), X \in [- \frac{1}{2}, \frac{1}{2}]$

Some More Questions From Inverse Trigonometric Functions Chapter

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

Phone Number

Email Address

Loaction

For Daily Free Study Material Join wiredfaculty

Inverse Trigonometric Functions

Prove that:3 sin-1 x = sin-1 (3x - 4x3), X ∈ -12, 12

Some More Questions From Inverse Trigonometric Functions Chapter

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

Prove that: $3 \sin^{- 1} x = \sin^{- 1} (3 x - 4 x^{3}), X \in [- \frac{1}{2}, \frac{1}{2}]$