Question

How do you prove sin(90°-a) = cos(a)?

Answer 1

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`:
`(a)+(90-a)+(90)=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 `sina`, which is defined as opposite (`h`) divided by hypotenuse (`s`) , equals `h/s` in the diagram:
`sina=h/s`

Note also that the cosine of the top angle, `90-a`, equals the adjacent side (`h`) divided by the hypotenuse (`s`):
`cos(90-a)=h/s`

So if `sina=h/s`, and `cos(90-a)=h/s`...

Then `sina` must equal `cos(90-a)`!
`sina=cos(90-a)`

And boom, proof complete.

Answer 2

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