Trig. Identity: 8sin^2(x/2)cos^2(x/2)=1-cos2x

[rebmik]

If anyone can point me in the right direction verifying this idenity I would really appreciate it. I am at a total loss.

8sin^2(x/2)cos^2(x/2)=1-cos2x

Thanks in advance for any help.

 

[soroban]

Re: Help -- Trouble w/ Trig. Identities

Hello, rebmik!

You're expected to be familiar with these identities:

. . $\displaystyle 2\sin\theta\cos\theta\:=\:\sin2\theta \;\;\;\;\sin^2\theta\:=\:\frac{1\,-\,\cos2\theta}{2}$

$\displaystyle 8\sin^2\left(\frac{x}{2}\right)\cos^2\left(\frac{x}{2}\right)\;=\;1\,-\,\cos2x$

We have: $\displaystyle \:8\sin^2\left(\frac{x}{2}\right)\cos^2\left(\frac{x}{2}\right)$

. . . . . $\displaystyle = \;2\left[4\sin^2\left(\frac{x}{2}\right)\cos^2\left(\frac{x}{2}\right)\right]$

. . . . . $\displaystyle = \;2\underbrace{\left[\sin\left(\frac{x}{2}\right)\cos\left(\frac{x}{2}\right)\right]}^2$

. . . . . $\displaystyle = \;\;\;\;\;2\overbrace{\left(\sin x\right)}$$\displaystyle ^2$

We have: $\displaystyle \:2\cdot\underbrace{\sin^2x}$

. . . . .$\displaystyle = \;2\overbrace{\left(\frac{1\,-\,\cos2x}{2}\right)}$

. . . . .$\displaystyle = \;1\,-\,\cos2x$

 

[rebmik]

Hi Soroban,

Thank you soooo much. I am familiar with those identities unfortunately, it just wasn't coming to me.

You make it look so easy!!!

Thanks again!