How do you prove sin(90°-a) = cos(a)?

I prefer a geometric proof. See below. 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.

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sin(90-x)=cos(x) - Trigonometry

Given: sin (A-B)=sin Acos B-cos Asin B. sin (90°)=1 cos (90°)=0