answer:I am assuming this isn’t homework, as it is summer and from what you wrote it appears you are merely curious, so I think it’s legal for me to do the proof here for you (don’t strike me down, Fluther gods!) So the limit definition of a derivative is lim h->0 (f(x+h)-f(x))/h. Let’s substitute the sin function in for f(x). limit h->0 (sin(x+h) – sin(x))/h A trigonometric identity says that sin(A + B) = sin(A)cos(B) + cos(A)sin(B) So let’s rewrite our limit using that: limit h->0 (sin(x)cos(h) + cos(x)sin(h) – sin(x))/h Rearrange that by removing common factors: limit h->0 (sin(x)(cos(h) – 1) + cos(x)sin(h))/h limit h->0 sin(x)(cos(h)-1)/h + cos(x)(sin(h)/h) The limit as h->0 of (cos(h)-1)/h is 0, and of sin(h)/h is 1 (You may already know why this is so I won’t prove it here, but if you do not know where I got that from, just let me know and I’ll be happy to do those proofs out for you too). Which leaves you with sin(x)(0) + cos(x)(1) or just cos(x). Cos is calculated very similarly; let me know if you’d like to see the proof for that as well. By toa do you mean tangent? I can prove that one too if you’d like.