<s>[INST] Find derivative $\frac{dy}{dx}$, given $y(x)=\sin^{-1}\left(\frac{5\sin x+4\cos x}{\sqrt{41}}\right)$ 
Find $\dfrac{dy}{dx}$ if $y=\sin^{-1}\bigg[\dfrac{5\sin x+4\cos x}{\sqrt{41}}\bigg]$

My Attempt
Put $\cos\theta=5/\sqrt{41}\implies\sin\theta=4/\sqrt{41}$
$$
y=\sin^{-1}\big[\sin(x+\theta)\big]\implies\sin y=\sin(x+\theta)\\
y=n\pi+(-1)^n(x+\theta)\\
\boxed{\frac{dy}{dx}=(-1)^n}
$$
But my reference gives the solution $y'=1$, am I missing something here ?
 [/INST] As you correctly pointed out we have
$$y = \arcsin\left[\sin\left(x +\theta\right) \right],$$
where $\theta = \arcsin\frac4{\sqrt{41}}$.
Now observe that ($k\in \Bbb Z$)
$$
\arcsin\sin \alpha = \begin{cases}\alpha-2k\pi & \left(2k\pi-\frac{\pi}2\leq \alpha < 2k\pi+\frac{\pi}2\right)\\
- \alpha - (2k-1)\pi & \left(2k\pi+\frac{\pi}2\leq \alpha < 2k\pi+\frac{3\pi}2\right).\end{cases}
$$
Thus your function is the triangular wave (see Figure below)
$$
y = 
\begin{cases}
x +\theta- 2k \pi & \left(2k\pi-\frac{\pi}2-\theta\leq x < 2k\pi+\frac{\pi}2-\theta\right)\\
-x-\theta-(2k-1)\pi-\theta & \left(2k\pi+\frac{\pi}2-\theta\leq x< 2k\pi+\frac{3\pi}2-\theta\right)
\end{cases}
$$
whose derivative is
$$
y = 
\begin{cases}
1 & \left(2k\pi-\frac{\pi}2-\theta< x < 2k\pi+\frac{\pi}2-\theta\right)\\
-1 & \left(2k\pi+\frac{\pi}2-\theta< x<2 k\pi+\frac{3\pi}2-\theta\right)
\end{cases}
$$

</s>