A bit of Calculus...

[qwertysaur]

OK this is more of an algebra problem but Derivatives makes it Calculus!

prove that the derivative f'(x) of

f(x) = sin(x) = cos(x)

I get the whole

(sin(x + h) - sin(x))/h but I have no idea what to do after that.

[Tavrobel]

If you're finding the derivative, then this definition comes with one important stipulation: the Limit as h --> 0.

What you can do here is use the trig identity that the sin(x*h) is sin(x)cos(h) + cos(x)sin(h). Go from there. The h on the bottom should drop out of the equation.

EDIT for solution: (SPOILER)If we use the limit definition of a function for a derivative as follows:

f'(x) = Lim h-->0 (f(x+h) - f(x))/h
f'(x) = Lim h-->0 (sin(x+h) - sin(x))/h

But we know that the sin function has a trigonometric identity. sin(a*b) = sin(a)cos(b) + cos(a)sin(b). The variables a and b can be substituted for the values, x and h.
sin(x*h) = sin(x)cos(h) + cos(x)sin(h)

Substitute that mess for sin(x*h).
f'(x) = Lim h-->0 (sin(x)cos(h) + cos(x)sin(h) - sin(x))/h
Special angles are possible to use, but you won't get rid of the h on the bottom, the effective 0. Factor out sin(x).
f'(x) = Lim h-->0 (sin(x)(cos(h) -1) + (cos(x)sin(h))/h

Now we can use some special angles. sin(h) = 0; cos(h) = 1, because h = 0.
f'(x) = Lim h-->0 (sin(x)(cos(h) -1) + (cos(x)sin(h))/h
Special limits. Lim x--> 0 (cos(x)-1)/x = 0; sin(x)/x = 1.
f'(x) = Lim h-->0 of 0 + (1)cos(x)
f'(sin(x)) = cos(x)

[qwertysaur]

thank you so much!

I knew I forgot some sort of trig property.