Question

Prove that

$\sf 3sin^{ - 1} x = sin ^{ - 1} (3x - 4 {x}^{3} )$
​

Answer 1

Answer:

this is the full explanation

Answer 2

Question:-

prove that $\sf 3sin^{ - 1} x = sin ^{ - 1} (3x - 4 {x}^{3} )$

Answer:-

firstly note that, this can be proven only if,

$\sf x \in \left[ - \dfrac{1}{2} , \dfrac{1}{2} \right]$

$\sf \: \therefore - \dfrac{1}{2} \leqslant x \leqslant \dfrac{1}{2} \:$

$\sf \implies \: - \dfrac{\pi}{6} \leqslant \theta \leqslant \dfrac{\pi}{6}$

multiply all three sides by 3,

$\sf - \dfrac{\pi}{6} \times 3\leqslant \theta \times 3 \leqslant \dfrac{\pi}{6} \times 3$

$\sf \implies \: - \dfrac{\pi}{2} \leqslant 3 \theta \leqslant \dfrac{\pi}{2}$

also, $\sf - \dfrac{1}{2} \leqslant x \leqslant \dfrac{1}{2} \:$

$\sf \implies -1 \leq 3x - 4x^3 \leq 1$

$\sf \therefore 3 sin \theta = 3x - 4x^3$

$\sf \implies 3\theta = \sin^{-1} (3x-4x^3)$

hence proven.

Additional Information:-

some important inverse identities

$0 \leqslant x \leqslant \pi \begin{cases} \sf tan^{-1} ( cot \: x) = \dfrac{\pi}{2} - x \\\\ \sf cot^{-1} ( tan \: x) = \dfrac{\pi}{2} - x \\\\ \sf sin^{-1} ( cos \: x) = \dfrac{\pi}{2} - x \\\\ \sf cos^{-1} ( sin \: x) = \dfrac{\pi}{2} - x \end{cases}$

$\rm$

$0 \leqslant x \leqslant \dfrac{\pi}{2} \begin{cases} \sf sec^{-1} ( cosec\: x) = \dfrac{\pi}{2} - x \\\\ \sf cosec^{-1} ( sec\:x) = \dfrac{\pi}{2} - x \end{cases}$

Inverse hyperbolic identities:~

$\sf (i) \;\;\; sinh^{-1} x = log_e (x + \sqrt{x^2 +1})$

$\sf (ii) \;\; cosh^{-1} x = log_e ( x + \sqrt{x^2-1})$

$\sf (iii) \; tanh^{-1} x = \dfrac{1}{2} \: log_e \left( \dfrac{1+x}{1-x} \right)$

$\sf (iv) \;\; coth^{-1} x = \dfrac{1}{2} \; log_e \left( \dfrac{x+1}{x-1} \right) , |x| > 1$

$\sf (v) \;\;\; sech^{-1} x = log_e \left( \dfrac{1+\sqrt{1-x^2}}{x} \right) , x \in (0,1]$

$\sf (vi) \;\; cosech^{-1} = \begin{cases} \sf log_e \left( \dfrac{1+\sqrt{1+x^2}}{x} \right), x > 0 \\\\ \sf log_e \left( \dfrac{1-\sqrt{1+x^2}}{x} \right), x < 0 \end{cases}$