R rebmik New member Joined Oct 26, 2006 Messages 4 Oct 26, 2006 #1 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.
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.
S soroban Elite Member Joined Jan 28, 2005 Messages 5,584 Oct 26, 2006 #2 Re: Help -- Trouble w/ Trig. Identities Hello, rebmik! You're expected to be familiar with these identities: . . \(\displaystyle 2\sin\theta\cos\theta\:=\:\sin2\theta \;\;\;\;\sin^2\theta\:=\:\frac{1\,-\,\cos2\theta}{2}\) \(\displaystyle 8\sin^2\left(\frac{x}{2}\right)\cos^2\left(\frac{x}{2}\right)\;=\;1\,-\,\cos2x\) Click to expand... 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: \(\displaystyle \:2\cdot\underbrace{\sin^2x}\) . . . . .\(\displaystyle = \;2\overbrace{\left(\frac{1\,-\,\cos2x}{2}\right)}\) . . . . .\(\displaystyle = \;1\,-\,\cos2x\)
Re: Help -- Trouble w/ Trig. Identities Hello, rebmik! You're expected to be familiar with these identities: . . \(\displaystyle 2\sin\theta\cos\theta\:=\:\sin2\theta \;\;\;\;\sin^2\theta\:=\:\frac{1\,-\,\cos2\theta}{2}\) \(\displaystyle 8\sin^2\left(\frac{x}{2}\right)\cos^2\left(\frac{x}{2}\right)\;=\;1\,-\,\cos2x\) Click to expand... 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: \(\displaystyle \:2\cdot\underbrace{\sin^2x}\) . . . . .\(\displaystyle = \;2\overbrace{\left(\frac{1\,-\,\cos2x}{2}\right)}\) . . . . .\(\displaystyle = \;1\,-\,\cos2x\)
R rebmik New member Joined Oct 26, 2006 Messages 4 Oct 26, 2006 #3 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!
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!
