# Differentiate $\sin^{-1}\left(\frac {\sin x + \cos x}{\sqrt{2}}\right)$ with respect to $x$

Differentiate $$\sin^{-1}\left(\frac {\sin x + \cos x}{\sqrt{2}}\right)$$ with respect to $x$.

I started like this: Consider $$\frac {\sin x + \cos x}{\sqrt{2}}$$, substitute $\cos x$ as $\sin \left(\frac {\pi}{2} - x\right)$, and proceed with the simplification. Finally I am getting it as $\cos \left(x - \frac {\pi}{4}\right)$. After this I could not proceed. Any help would be appreciated. Thanks in advance!

• @user222031 I also tried like that but I ended up with $\frac {\cos x - \sinx}{\sqrt (1 - \sin 2x)}$. How to proceed after this? May 5, 2015 at 13:28
• $\sin x \cos \pi/4 + \cos x \sin \pi/4$
– Mann
May 5, 2015 at 13:29
• @mann Where to use this? Sorry I don't get it May 5, 2015 at 13:30
• Look the answers below, you'd see it.
– Mann
May 5, 2015 at 13:31

Hint The angle sum rule for $\sin$ is $$\sin(x + y) = \sin x \cos y + \cos x \sin y.$$

Additional hint In particular, if we take $y = \frac{\pi}{4}$ and rearrange, we get $$\sin x + \cos x = \sqrt{2} \sin\left(x + \frac{\pi}{4}\right).$$

It's better to rewrite $$\frac{1}{\sqrt{2}}(\sin x+\cos x)=\sin(x+\pi/4)$$ and then use the chain rule: $$f(x)=\arcsin\sin\Bigl(x+\frac{\pi}{4}\Bigr)$$ so $$f'(x)=\frac{1}{\sqrt{1-\sin^2(x+\pi/4)}}\cos\Bigl(x+\frac{\pi}{4}\Bigr) =\dots$$ (Beware of the square root!)

$f'(x)=\dfrac{\cos(x+\pi/4)}{|\cos(x+\pi/4)|}$

so the derivative is $1$ where $\cos(x+\pi/4)>0$ and $-1$ where $\cos(x+\pi/4)<0$; the function is not differentiable where $\cos(x+\pi/4)=0$.

• Then $\sin^{-1} \Sin (x + \frac {\pi}{4})$ is simply $x + \frac {\pi}{4}$ and if we differentiate it we will get 1. Am I right? May 5, 2015 at 13:33
• @Sahana The equality $\arcsin\sin\alpha=\alpha$ is generally false. For instance, $\arcsin\sin\pi=\arcsin 0=0\ne\pi$. May 5, 2015 at 13:34
• BEWARE OF THE SQUARE ROOT.
– Mann
May 5, 2015 at 13:35
• If I simplify @egreg expression also the final ans is coming as $1$. So I think in this case $\sin^{-1}(\sin x) = x$. is it ok? May 5, 2015 at 13:38
• Still no. Greg final expression is not always 1.
– Mann
May 5, 2015 at 13:40

An alternative approach is to use Implicit Differentiation:

$$$$y = \arcsin\left(\frac{\sin(x) + \cos(x)}{\sqrt{2}} \right) \rightarrow \sin(y) = \frac{\sin(x) + \cos(x)}{\sqrt{2}}$$$$

Now differentiate with respect to '$$x$$':

\begin{align} \frac{d}{dx}\left[\sin(y) \right] &= \frac{d}{dx}\left[\frac{\sin(x) + \cos(x)}{\sqrt{2}} \right] \\ \cos(y)\frac{dy}{dx} &= \frac{\cos(x) - \sin(x)}{\sqrt{2}} \\ \frac{dy}{dx} &= \frac{\cos(x) - \sin(x)}{\sqrt{2}\cos(y)} \end{align}

Thus:

$$$$\frac{dy}{dx} = \frac{d}{dx}\left[\arcsin\left(\frac{\sin(x) + \cos(x)}{\sqrt{2}} \right) \right] = \frac{\cos(x) - \sin(x)}{\sqrt{2}\cos\left(\arcsin\left(\frac{\sin(x) + \cos(x)}{\sqrt{2}}\right) \right)}$$$$

Here this method is unnecessarily complicated in comparison to those already presented. It is however good to know if an identity is either unknown.

$$\frac{d(\arcsin x)}{dx}=\frac{1}{\sqrt{1-x^2}}$$ It is as simple as that. This is followed by the substitution $$v(x)=\frac{\sin x+\cos x}{\sqrt{2}}$$. \begin{align} \frac{d}{dx}\arcsin(v(x)) & =\frac{1}{\sqrt{1-v^2(x)}}v'(x) \\ &= \frac{1}{\sqrt{1-\Biggl(\frac{\sin x+\cos x}{\sqrt{2}}\Biggl)^2}}\Biggl(\frac{\cos x-\sin x}{\sqrt{2}}\Biggl) \\ &= \frac{\sqrt{2}}{\sqrt{(\cos x-\sin x)^2}}\frac{\cos x-\sin x}{\sqrt{2}} \\ &= \frac{\cos x-\sin x}{|\cos x-\sin x|} \\ &= \frac{\cos(x+\frac{\pi}{4})}{|\cos(x+\frac{\pi}{4})|} \\ \end{align}