We know Sin 2x - Sin 2y = 2 cos(x+y) sin(x-y) (The Trig Equiv page has a proof of this)
So it follows that sin(x+h) - sin(x) = 2 cos(x+h/2) sin(h/2)
If you divide both sides by h, and take the limit as h approaches zero, you get
d/dx (sin x) = cos(x) limh−>0 sin(h/2)/(h/2)
That limit is 1, of course (proof) so the result follows.
Proof that lim(x−>0) sin x / x = 1
Sin or Cos 3x, 4x, etc. -- trig functions of any multiple of an angle.
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