1. ## cos2x

How can i prove cos2x derivation using cos x by basic derivation proof.... like (limit and delta(x) delta(y))

2. ## Re: cos2x

just expand out $\frac{\cos\left(2(x+h)\right)-\cos(2x)}{h}$ and take the limit as $h \to 0$.

3. ## Re: cos2x

but i asked to use cos2x proofing by cosx... i tried this way,,, cos2x=2cos^2(x)-1, and then expand, finally i stacked up.... please give me a solution????

4. ## Re: cos2x

Originally Posted by srirahulan
but i asked to use cos2x proofing by cosx... i tried this way,,, cos2x=2cos^2(x)-1, and then expand, finally i stacked up.... please give me a solution????
sheesh

$\frac{\cos\left(2(x+h)\right)-\cos(2x)}{h}=$

$\frac{\cos(2x+2h)-cos(2x)}{h}=-2\sin(2x)$

$\frac{\cos(2x)\cos(2h)-\sin(2x)\sin(2h)-\cos(2x)}{h}=$

$\frac{\cos(2x)(cos(2h)-1)-\sin(2x)\sin(2h)}{h}=$

$\cos(2x)\frac{cos(2h)-1}{h}-\sin(2x)\frac{\sin(2h)}{h}$

we need to reduce the sin and cos terms that involve h further so we can use known limits

$\cos(2h)=2\cos^2(h)-1 and \sin(2h)=2\sin(h)\cos(h)$

so

$\displaystyle{\lim_{h\to 0}}\frac{\cos(2h)-1}{h}=\displaystyle{\lim_{h\to 0}}\frac{2\cos^2(h)-2}{h}=0$

$\displaystyle{\lim_{h\to 0}}\frac{\sin(2h)}{h}=\displaystyle{\lim_{h\to 0}}\frac{2\sin(h)\cos(h)}{h}=2$

and thus

$\displaystyle{\lim_{h\to 0}}\frac{\cos\left(2(x+h)\right)-\cos(2x)}{h}=-2\sin(2x)$

5. ## Re: cos2x

Originally Posted by srirahulan
How can i prove cos2x derivation using cos x by basic derivation proof.... like (limit and delta(x) delta(y))
It depends on what you already know. Are you allowed to use the Chain Rule? If so, you can prove \displaystyle \begin{align*} \frac{d}{dx} \left[ \cos{(x)} \right] = -\sin{(x)} \end{align*} by first principles and then apply the chain rule. If you really wanted to, you could prove \displaystyle \begin{align*} \frac{d}{dx} \left[ \sin{(x)} \right] = \cos{(x)} \end{align*} by first principles, then remember \displaystyle \begin{align*} \cos{(x)} \equiv \sin{ \left( \frac{\pi}{2} - x \right) } \end{align*} and use the chain rule from there...