# Def. of derivative and cosx=sin(Pi/2-x) to prove y'=-sinx

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1. Jan 5, 2016

### ThaSlave

A lot of web pages/books show how to use cosx=sin(Pi/2-x) and the chain rule to prove that the derivative of
cosx=-sinx. My question is how to use this identity and the defintion of the derivative to prove the same thing.
Or whether it is at all possible. Seeing that i get dy/dx=(cos(x+h)-cosx)/h = (sin(Pi/2-x-h)-sin(Pi/2-x))/h =
(sin(Pi/2-x)cosh-cos(pi/2-x)sinh-sin(pi/2-x))/h. The first term here is the problem, namely sin(Pi/2-x)cosh/h, since this needs to be equal to sin(Pi/2-x) for this to work. Is there another way around?

Another way of thinking is to say that cosh goes to 1 as h goes to 0, but using the same reasoning with sinh leaves me with dy/dx=0.

2. Jan 5, 2016

### blue_leaf77

Just continue from there. The way you proceed is to express $\cos h$ and $\sin h$ in terms of their respective Taylor expansion. Let's see what you get after doing this. For example, substituting the expansion for $\cos h$ to the first term leads to
$$\sin(\pi/2-x)\cos h = \sin(\pi/2-x)\left(1-\frac{h^2}{2!}+\frac{h^4}{4!} - \ldots \right)$$
Now substract the 3rd term $\sin(\pi/2-x)$ from the RHS of the above equation.

3. Jan 5, 2016

### Staff: Mentor

Rather than using the identity you're trying to use, just continue from the above.
$\frac{\cos(x + h) - \cos(x)}{h} = \frac{\cos(x)\cos(h) - \sin(x)\sin(h) - \cos(x)}{h} = \frac{\cos(x)(\cos(h) - 1)}{h} - \frac{\sin(x) \sin(h)}{h}$.
To get the derivative of cos(x), take the limits of the two expressions above.