Question

Verify the following identity used in calculus:
(cos(x+h)-cosx)/h = (cosx(cosh-1))/h = (sinxsinh)/h

Answer 1

\[\frac{\cos(x+h)-\cos x}{h}\]
Angle sum identity for cosine:
\[\cos(x+h)=\cos x\cos h-\sin x\sin h\]
\[\begin{align*}\frac{\cos(x+h)-\cos x}{h}&=\frac{\cos x\cos h-\sin x\sin h-\cos x}{h}\\
&=\frac{\cos x(\cos h-1)-\sin x\sin h}{h}
\end{align*}\]

Answer 2

I did the same thing as you did but the equation has \[\frac{ cosx(\cosh-1) }{ h } = \frac{ sinxsinh }{ h }\] which is not what you got. It's like they split it in half? I don't understand...

Answer 3

Could be that the last fraction I have is set equal to 0, or a typo. In any case, what we both have is correct.

It looks to me like you're being slowly introduced to the definition of the derivative. Namely,
\[\frac{d}{dx}\cos x=\lim_{h\to0}\frac{\cos(x+h)-\cos x}{h}\]
Then after some rewriting (see my last post), you have
\[\begin{align*}\frac{d}{dx}\cos x&=\lim_{h\to0}\frac{\cos x(\cos h-1)-\sin x\sin h}{h}\\
&=\lim_{h\to0}\frac{\cos x(\cos h-1)}{h}-\lim_{h\to0}\frac{\sin x\sin h}{h}\\
&=\cos x\left(\lim_{h\to0}\frac{\cos h-1}{h}\right)-\sin x\left(\lim_{h\to0}\frac{\sin h}{h}\right)
\end{align*}\]
Those last two limits are
\[\lim_{h\to0}\frac{\cos h-1}{h}=0~\text{ and }~\lim_{h\to0}\frac{\sin h}{h}=1\]
So you have \(\dfrac{d}{dx}\cos x=-\sin x\).

The "=" that you say is there must be a mistake, because it doesn't make sense.