anonymous
  • anonymous
Use the squared Identities to simplify 2cos^2 (x) * cos^2 (x)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
@ganeshie8 @JackJordan
anonymous
  • anonymous
uhmm... how simple do you want it I mean an easy one would be just 2cos^4(x) lol
anonymous
  • anonymous
a) 1-cos(4x) / 4 b) 1+4cos(2x)+cos(4x) / 4 c) 1+cos(4x) / 4 d) 1+4cos(2x)-cos(4x) / 4

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More answers

anonymous
  • anonymous
@ganeshie8
anonymous
  • anonymous
are you sure you haven't miswritten the second one?
anonymous
  • anonymous
b) 1+4cos(2x)+cos(4x) / 4
anonymous
  • anonymous
thats what i have
anonymous
  • anonymous
@ganeshie8
anonymous
  • anonymous
@JackJordan
anonymous
  • anonymous
@jim_thompson5910
anonymous
  • anonymous
someone help!
hartnn
  • hartnn
write both \(\cos^2 x = \dfrac{1-\cos 2x}{2}\)
anonymous
  • anonymous
It must be a typo... I'm getting 1+4cos(2x)+cos(4x) + 3 / 4 sorry.
hartnn
  • hartnn
***\(\cos^2 x = \dfrac{1+\cos 2x}{2}\)
anonymous
  • anonymous
elementary mistake hartnn
hartnn
  • hartnn
and then expand it.
anonymous
  • anonymous
So my attempt leads to something similar to B but with +3 at the end. Either I'm wrong (not likely) or the question is wrong.
hartnn
  • hartnn
then using the same identity you'll need to write \(\Large \cos^2 2x = \dfrac{1+\cos 4x}{2}\)
anonymous
  • anonymous
ROOKIE MISTAKE!!!
anonymous
  • anonymous
sorry to everyone. I am so sorry
anonymous
  • anonymous
So I was right
hartnn
  • hartnn
so you got it now?
anonymous
  • anonymous
i must've made a huge typo
anonymous
  • anonymous
the question is 2sin^2 (x) cos^2x
anonymous
  • anonymous
sin. not cos. i can't believe i made that mistake.
hartnn
  • hartnn
okk, then write \(\large \sin^2 x = \dfrac{1-\cos 2x}{2}\) thats the only difference
anonymous
  • anonymous
Dang it I knew you made that mistake! Lol, it gives a difference of two squares at the top. I already did that. The answer is a. Can you see why?
hartnn
  • hartnn
I'd like to see sbsbrand's attempt! :)
anonymous
  • anonymous
yay it worked! thanks everyone that stayed to help me even tho i wrote the question wrong.......XD

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