First slide

Introduction to ITF

Question

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

Moderate

A

$3 \sin^{- 1} x$
B

$- π + 3 \sin^{- 1} x$
C

$- (π + 3 \sin^{- 1} x)$
D

$π - 3 \sin^{- 1} x$

Solution

$L e t x = \sin θ . S i n c e x \in [\frac{1}{2}, 1], \Rightarrow \sin θ \in [\frac{1}{2}, 1] \Rightarrow \frac{π}{6} \leq θ \leq \frac{π}{2} \Rightarrow \frac{π}{2} \leq 3 θ \leq \frac{3 π}{2} \Rightarrow - \frac{3 π}{2} \leq - 3 θ \leq - \frac{π}{2} \Rightarrow π - \frac{3 π}{2} \leq π - 3 θ \leq π - \frac{π}{2} \Rightarrow - \frac{π}{2} \leq π - 3 θ \leq \frac{π}{2} N o w \sin^{- 1} (3 x - 4 x^{3}) = \sin^{- 1} (3 \sin θ - 4 \sin^{3} θ) = \sin^{- 1} (\sin 3 θ) = \sin^{- 1} (\sin (π - 3 θ)) = π - 3 θ = π - 3 \sin^{- 1} x$

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