How do you prove sin(90°-a) = cos(a)?

2 Answers
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:
enter image source here
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.
enter image source here
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.

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