Find the area between the curves $y = sin (x)$ and $y = cos(x)$

There was this post from one of my junior fellow from high school who is enthusiastic on mathematics, physics and electronics. Here’s his post:

I just wanted to unrust my mind. And hence this:

First, the point where two curves $y = sin(x)$ and $y = sin(x)$ meet are those points where their $y$ values are same. To find that,

$sin(x) = cos(x)$,

Or, $sin(x) - cos(x) = 0$,

Old school math says to multiply it by $\frac{1}{\sqrt{2}}$ on both sides, that gives:

$\frac{1}{\sqrt{2}} sin (x) - \frac{1}{\sqrt{2}} cos(x) = 0$,

Or, $cos(\frac{\pi}{4})sin(x) - sin(\frac{\pi}{4})cos(x) = 0$

Or, $sin(x - \frac{\pi}{4}) = 0$

If $sin(\theta) = 0$, then $\theta = \pm n \pi$, where $n=0, 1, 2, 3, …$

So, in our case $x - \frac{\pi}{4} = 0$ and $x - \frac{\pi}{4} = 1$ gives relevant solution.

i.e. $x_1 = \frac{\pi}{4}$ and $x_2 = \frac{5\pi}{4}$.

Now that we know the limits of the integral, we have to find the area of the upper curve viz. $y = sin(x)$ and subtract the area of the lower curve viz. $y = cos(x)$ from the upper one.

i.e. Required Area $(A) = \int_{\frac{\pi}{4}}^{\frac{5\pi}{4}}sin(x)dx - \int_{\frac{\pi}{4}}^{\frac{5\pi}{4}}cos(x)dx$

Or, $A = [- cos(x) - sin (x)]_{\frac{\pi}{4}}^{\frac{5\pi}{4}}$

Or, $A = [-(-\frac{1}{\sqrt{2}}) - (-\frac{1}{\sqrt{2}})] - [- \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}] $

$\therefore A = \frac{4}{\sqrt(2)} = 2.828 $ in square units.

So, that’s it! That’s the area!

Last Updated: Thursday, 5 Mar, 2020, 21:41 NPT
Author: Madhav Humagain (scimad)