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

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

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

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

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

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

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

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

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

so

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

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

and thus

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

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 \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 \displaystyle \begin{align*} \frac{d}{dx} \left[ \sin{(x)} \right] = \cos{(x)} \end{align*} by first principles, then remember \displaystyle \displaystyle \begin{align*} \cos{(x)} \equiv \sin{ \left( \frac{\pi}{2} - x \right) } \end{align*} and use the chain rule from there...