Derivative of trigonometric functions
When we have trigonometric functions, the derivatives are all related to the trigonometric identities. Let's plot sine, cosine and tangent on the same space:
The rate of change of the sine and the tangent invert its sign at the origin and also coincides with the root. The point where the sine is maximum is also the point where the cosine is zero. The opposite is also true, where the cosine is maximum the sine is zero. The point where the cosine and the sine intercept each other is the angle [math]\displaystyle{ \pi / 4 }[/math] in the unit circle. The point where the tangent and the cosine intercept each other is the angle [math]\displaystyle{ 3 \pi / 4 }[/math].
Sine and cosine are identical functions with the exception that their respective roots differ by an angle of [math]\displaystyle{ \pi / 2 }[/math]. We have that [math]\displaystyle{ \sin(x + \pi / 2) = \cos(x) }[/math] and [math]\displaystyle{ cos(x - \pi / 2) = \sin(x) }[/math]. The fact that both functions can match each other is what we need to prove that cosine is the derivative of sine.
[math]\displaystyle{ \lim_{h \ \to \ 0} \sin(x) = \frac{\sin(x + h) - \sin(x)}{h} }[/math]
The formula for the sum of angles doesn't help because it's going to complicate the problem by adding a second variable that we don't need. What we need is an identity that transforms a difference of sines into a product and there is such identity. The idea is that sine and cosine "disappear" at certain angles, because we end up multiplicating by zero or one depending on the angle:
[math]\displaystyle{ \lim_{h \ \to \ 0} \sin(x) = \frac{2 \sin \left(\frac{h}{2}\right) \cos \left(\frac{2x + h}{2}\right)}{h} }[/math] (the limit of the product is the product of the limits)
[math]\displaystyle{ \lim_{h \ \to \ 0} \sin(x) = 2 \lim_{h \ \to \ 0} \frac{\sin(h/2)}{h/2} \lim_{h \ \to \ 0} \cos \left(\frac{2x + h}{2}\right) }[/math] (we can move a constant factor out of the limit and multiply the limit by the inverse of that constant to keep the same result)
[math]\displaystyle{ \lim_{h \ \to \ 0} \sin(x) = 1 \cos(x) }[/math] (the fundamental trigonometric limit is equal to one)
