# cos2x

• Jan 6th 2014, 07:49 PM
srirahulan
cos2x
How can i prove cos2x derivation using cos x by basic derivation proof.... like (limit and delta(x) delta(y))
• Jan 6th 2014, 07:58 PM
romsek
Re: cos2x
just expand out $\displaystyle \frac{\cos\left(2(x+h)\right)-\cos(2x)}{h}$ and take the limit as $\displaystyle h \to 0$.
• Jan 6th 2014, 08:16 PM
srirahulan
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????
• Jan 6th 2014, 09:14 PM
romsek
Re: cos2x
Quote:

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

$\displaystyle \displaystyle{\lim_{h\to 0}}\frac{\cos\left(2(x+h)\right)-\cos(2x)}{h}=-2\sin(2x)$
• Jan 6th 2014, 09:48 PM
Prove It
Re: cos2x
Quote:

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...
• Jan 13th 2014, 08:18 PM
ibdutt
Re: cos2x