The angle between the curves $y=\sin x\text{ and }y=\cos x$

$0<x<\frac{\pi }{2}$, is

Solution:

Curves $y = \sin x$ and $y = \cos x$ intersect at

$P(\pi /4,1\sqrt{2})$

Now,

$\begin{array}{l}y=\sin x\Rightarrow \frac{dy}{dx}=\cos x\Rightarrow m_1={\left(\frac{dy}{dx}\right)}_P=\cos \frac{\pi }{4}=-\frac{1}{\sqrt{2}}\\ y=\cos x\Rightarrow \frac{dy}{dx}=-\sin x\Rightarrow m_2={\left(\frac{dy}{dx}\right)}_{\rho }=-\sin \frac{\pi }{4}=-\frac{1}{\sqrt{2}}\end{array}$

Let $\theta$ be the angle between the given curves at $P$. Then, 

$\begin{array}{l}\tan \theta =\frac{m_1-m_2}{1+m_1{m}_2}\\ \Rightarrow \tan \theta =\frac{\sqrt{2}}{1-\frac{1}{2}}=2\sqrt{2}\Rightarrow 0={\tan }^{-1}\left(2\sqrt{2}\right)\end{array}$