# How do you verify sin^2 x + sin^2(pi/2 -x) = 1?

May 15, 2015

On the trig unit circle:

$\sin \left(\frac{\pi}{2} - x\right) = \cos x$, then

${\sin}^{2} x + {\sin}^{2} \left(\frac{\pi}{2} - x\right) = {\sin}^{2} x + {\cos}^{2} x = 1$

May 17, 2015

If $0 < x < \frac{\pi}{2}$ then another way to picture this is as follows:

Construct a right angled triangle with angles $x$, $\left(\frac{\pi}{2} - x\right)$ and $\frac{\pi}{2}$, with hypotenuse of length $1$. It is possible to do this since these angles add up to $\pi$.

Then the side opposite the angle $x$ will have length $\sin x$ and the side opposite the angle $\left(\frac{\pi}{2} - x\right)$ will have length $\sin \left(\frac{\pi}{2} - x\right)$.

By Pythagoras theorem, the sum of the squares of the lengths of these sides is equal to the square of the length of the hypotenuse.

So ${\sin}^{2} x + {\sin}^{2} \left(\frac{\pi}{2} - x\right) = {1}^{2} = 1$