The trigonometric equation sin ^{-1} x = 2 si | vedclass

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Question

(C) The equation $\sin ^{-1} x = 2 \sin ^{-1} a$ has a solution for $x$ if and only if the right-hand side $2 \sin ^{-1} a$ lies within the range of t

Vedclass · Accepted Answer

only when $|a| \leqslant \frac{1}{\sqrt{2}}$

Vedclass · Answer

(C) The equation $\sin ^{-1} x = 2 \sin ^{-1} a$ has a solution for $x$ if and only if the right-hand side $2 \sin ^{-1} a$ lies within the range of the $\sin ^{-1}$ function,which is $[-\frac{\pi}{2}, \frac{\pi}{2}]$.Thus,we must have $-\frac{\pi}{2} \leq 2 \sin ^{-1} a \leq \frac{\pi}{2}$.Dividing by $2$,we get $-\frac{\pi}{4} \leq \sin ^{-1} a \leq \frac{\pi}{4}$.Applying the sine function to all parts,since $\sin$ is an increasing function,we get $\sin(-\frac{\pi}{4}) \leq a \leq \sin(\frac{\pi}{4})$.This simplifies to $-\frac{1}{\sqrt{2}} \leq a \leq \frac{1}{\sqrt{2}}$,which is equivalent to $|a| \leq \frac{1}{\sqrt{2}}$.

The trigonometric equation $\sin ^{-1} x = 2 \sin ^{-1} a$ has a solution

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