# Ask Uncle Colin: The Area In Between

Dear Uncle Colin,

I have the graphs of $y=\sin(x)$ and $y=\cos(x)$ for $0 < x < 2\pi$. They cross in two places, and I need to find the area enclosed.

I’ve figured out that they cross at $\piby 4$ and $\frac{5}{4}\pi$, but after that I’m stuck!

- Probably A Simple Calculation, Absolutely Lost

Hi, PASCAL, and thanks for your message!

You’ll probably be unsurprised to learn there are several ways to do it.

### Directly

$\int_{\piby4}^{\frac{5}{4}\pi} \sin(x) - \cos(x) \dx = \left[ -\cos(x) - \sin(x) \right]_{\piby4}^{\frac{5}{4}\pi}$

$\dots = \left[ \left( \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} \right) - \left( - \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}\right) \right]$

$\dots = 2\sqrt{2}$.

### Harmonically

$\sin(x) - \cos(x) \equiv \sqrt{2} \sin\left( x - \piby 4\right)$

$\int_{\piby4}^{\frac{5}{4}\pi} \sin(x) - \cos(x) \dx =\sqrt{2} \int_{0}^{\pi} \sin(t) \dt$ (after a sneaky variable change).

$\dots = \sqrt{2} \left[ -\cos(\pi) - \cos(0) \right]$

$\dots = 2\sqrt{2}$

I can’t help but feel there’s an even nicer way, though!

Hope that helps,

- Uncle Colin