# Solve for $x$ in $\cos(2 \sin ^{-1}(- x)) = 0$

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?

• You didn't take multiple values into account. Neither cos nor sin are one to to one. – fleablood Jun 8 '17 at 15:34

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 $\displaystyle \boxed{\pm \frac{1}{\sqrt{2}}}$.

• Because cos^-1 only returns one value. You have to use symmetry to get the other value – Saketh Malyala Jun 8 '17 at 15:43
• How did you get This should give you $(1-(-x)^2)-(-x)^2=0$. – Neha Gupta Jun 8 '17 at 15:49
• remember $\cos(\sin^{-1}(x)) = \sqrt{1-x^2}$ by right angle trigonometry – Saketh Malyala Jun 8 '17 at 15:50
• $\sin(\sin^{-1}(x))=x$ because sin is the inverse of $\sin^{-1}$ – Saketh Malyala Jun 8 '17 at 15:51

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?

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 $x = \mp \frac {1}{\sqrt{2}}$

Let $\sin^{-1}(-x)=u\implies\sin u=-x$

$$\cos(2\sin^{-1}(-x))=\cos2u=1-2\sin^2u=1-2(-x)^2$$