The angle at which the curves $y=\sin x$ and $y=\cos x$ intersect in $[0, \pi]$ is

$\pm \tan ^{-1} 2 \sqrt{2}$

Clearly, curves $y=\sin x$ and $y=\cos x$ intersect at $x=\frac{\pi}{4}$ in $[0, \pi]$. Thus, the coordinates of the point P of intersection of the two curves are $P(\pi / 4,1 / \sqrt{2})$

Clearly,

$m_1=\left(\frac{d y}{d x}\right)_P=\cos \frac{\pi}{4}=\frac{1}{\sqrt{2}}$ for the curve $y=\sin x$

and,

$m_2=\left(\frac{d y}{d x}\right)_P=-\sin \frac{\pi}{4}=-\frac{1}{\sqrt{2}}$ for the curve $y=\cos x$

Let $\theta$ be the angle of intersection of the two curves at point P.

Then,

$\tan \theta=\left|\frac{m_1-m_2}{1+m_1 m_2}\right|$

$\Rightarrow \tan \theta=\left|\frac{\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}}{1-\frac{1}{2}}\right|=2 \sqrt{2}$

$\Rightarrow \quad \theta=\tan ^{-1}( \pm 2 \sqrt{2})= \pm \tan ^{-1}(2 \sqrt{2})$