Solve $$\cos(2 \sin ^{-1}(- x)) = 0$$
I get the answer $\frac{-1}{ \sqrt2}$
By solving like this
\begin{align}2\sin ^{-1} (-x )&= \cos ^{-1} 0\\ 2\sin^{-1}(- x) &= \frac\pi2\\ \sin^{-1}(- x) &= \frac\pi4\\ -x &= \sin\left(\frac\pi4\right)\end{align}
Thus $x =\frac{-1}{\sqrt2}$
But correct answer is $\pm\frac{1}{\sqrt2}$
Where am I going wrong?
$\endgroup$ 14 Answers
$\begingroup$Okay so expand the double angle to get $\cos^2(\sin^{-1}(-x))-\sin^2(\sin^{-1}(-x)=0$.
This should give you $(1-(-x)^2)-(-x)^2=0$.
Finally you have $1-2x^2=0$.
Solutions are $\displaystyle \boxed{\pm \frac{1}{\sqrt{2}}}$.
$\endgroup$ 4 $\begingroup$You need to solve $\cos \left(2 \arcsin(-x) \right) = 0$. Let $y = 2 \arcsin(-x)$ then $\cos y = 0$ so $y = \pi/2 \pm n\pi$. Then, $$ 2 \arcsin(-x) = \frac{pi}{2} \pm n\pi $$ which implies $$ x = -\sin \left( \frac{\pi}{4} \pm \frac{n\pi}{2} \right) $$
Can you simplify this?
$\endgroup$ $\begingroup$Trig functions are not one-to-one. For example if $\sin x = \frac 12$ then $x$ might be equal to $\frac {\pi}6$, but $x$ could also be $\frac {5\pi}6$ or $\frac {13\pi}6$ or it could be any $2k\pi + \frac {\pi}6$ or $(2k+1)\pi - \frac {\pi} 6$.
So when we say $\sin^{-1} x = \theta$ we are not just saying that $\theta$ is the angle so that $\sin (\theta) = x$. There should be an infinite number of such angles. We are saying that $\theta$ is an angle so that $\sin (\theta) = x$ AND we are saying that $\frac {-\pi}2 \le \theta \le \frac {\pi}2$. For those two conditions there is only one possible theta.
So $\sin^{-1}(\sin x)$ may or MAY NOT be equal to $x$. For example $\sin^{-1}(\sin (\frac {5\pi}6)) = \sin^{-1}(\frac 12) = \frac {\pi}6 \ne \frac {5\pi}6$.
However $\sin^{-1}(\sin x) = x + 2k\pi$ for some integer $k$ or $\sin^{-1}(\sin x) = (2k+ 1)\pi - x$.
So 1) Because $\sin (x) = \sin (\pi - x)$ and $x = \frac {\pi}2 - (\frac {\pi}2 - x)$ and $\pi - x = \frac{\pi}2 + (\frac {\pi}2 - x)$ then $\sin^{-1}(x) = (\frac {\pi}2 \pm x) \pm 2k\pi)$.
$\frac {-\pi}2 \le sin^{-1}(x) \le \frac {\pi}2$
And 2) Because $\cos (x) = \cos(\pi+ x)$ we have $\cos^{-1}(\cos x) = x \pm k \pi$.
$0 \le \cos^{-1} (x) \le \pi$
So ....
$\cos (2\sin^{-1}(-x) ) = 0$
$-\frac {\pi}2 \le \sin^{-1}(-x) \le \frac {\pi}2$ so $-\pi \le 2\sin^{-1}(-x) \le \pi$ and $\sin^{-1}(-x) = \cos^{-1}(0) \pm k\pi = \frac {\pi}2 \pm k\pi$. So $2\sin^{-1}(-x) = \pm \frac {\pi}2$
And $\sin^{-1}(-x) = \pm \frac {\pi}4$
So $\sin (\pm \frac {\pi} 2) = -x$
So $\pm \frac {1}{\sqrt{2}} = -x$
So $x = \mp \frac {1}{\sqrt{2}}$
$\endgroup$ $\begingroup$Let $\sin^{-1}(-x)=u\implies\sin u=-x$
$$\cos(2\sin^{-1}(-x))=\cos2u=1-2\sin^2u=1-2(-x)^2$$
$\endgroup$
