# How do you find the exact value of the area in the first quadrant enclosed by graph of y=sinx and y=cosx?

Jan 19, 2017

$2 \sqrt{2} - 2$

#### Explanation:

The area enclosed by the graphs $y = \sin x$ and $y = \cos x$ is shaded in the above graph.

By symmetry this area is $2 A$, where

$A = {\int}_{0}^{\frac{\pi}{4}} \cos x - \sin x \setminus \mathrm{dx}$
$\setminus \setminus \setminus = {\left[\sin x + \cos x\right]}_{0}^{\frac{\pi}{4}}$
$\setminus \setminus \setminus = \left\{\left(\sin \left(\frac{\pi}{4}\right) + \cos \left(\frac{\pi}{4}\right)\right) - \left(\sin 0 + \cos 0\right)\right\}$
$\setminus \setminus \setminus = \left(\frac{1}{2} \sqrt{2} + \frac{1}{2} \sqrt{2}\right) - \left(0 + 1\right)$
$\setminus \setminus \setminus = \sqrt{2} - 1$

Hence shaded are is $2 A = 2 \sqrt{2} - 2$