# How do you prove sin(90°-a) = cos(a)?

Jun 30, 2016

I prefer a geometric proof. See below.

#### Explanation:

If you're looking for a rigorous proof, I'm sorry - I'm not good at those. I'm sure another Socratic contributor like George C. could do something a little more solid than I can; I'm just going to give the lowdown on why this identity works.

Take a look at the diagram below:

It's a generic right triangle, with a ${90}^{o}$ angle as indicated by the little box and an acute angle $a$. We know the angles in a right triangle, and a triangle in general, must add to ${180}^{o}$, so if we have an angle of $90$ and an angle of $a$, our other angle must be $90 - a$:
$\left(a\right) + \left(90 - a\right) + \left(90\right) = 180$
$180 = 180$

We can see that the angles in our triangle do indeed add to $180$, so we're on the right track.

Now, let's add some variables for side length onto our triangle.

The variable $s$ stands for the hypotenuse, $l$ stands for length, and $h$ stands for height.

We can start on the juicy part now: the proof.

Note that $\sin a$, which is defined as opposite ($h$) divided by hypotenuse ($s$) , equals $\frac{h}{s}$ in the diagram:
$\sin a = \frac{h}{s}$

Note also that the cosine of the top angle, $90 - a$, equals the adjacent side ($h$) divided by the hypotenuse ($s$):
$\cos \left(90 - a\right) = \frac{h}{s}$

So if $\sin a = \frac{h}{s}$, and $\cos \left(90 - a\right) = \frac{h}{s}$...

Then $\sin a$ must equal $\cos \left(90 - a\right)$!
$\sin a = \cos \left(90 - a\right)$

And boom, proof complete.

Jun 30, 2016

sin (90 - a) = cos a

#### Explanation:

Another way is to apply the trig identity:
sin (a - b) = sin a.cos b - sin b.cos a
sin (90 - a) = sin 90.cos a - sin a cos 90.
Since sin 90 = 1, and cos 90 = 0, therefor,
sin (90 - a) = cos a