anonymous
  • anonymous
Prove: 32(sin^4x)(cos^2x)=2-cos2x-2cos4x+cos6x Please help :(
Precalculus
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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.Sam.
  • .Sam.
The RHS has all the cosine function, you might wanna change the LHS sine function to cosine
.Sam.
  • .Sam.
\[32(\sin^4x)(\cos^2x)\] \[\Rightarrow32(\sin^2(x)\sin^2(x))(\cos^2x)\] Using \[\sin^2x=1-\cos^2x\] \[\Rightarrow32((1-\cos^2x)(1-\cos^2x))(\cos^2x)\]
.Sam.
  • .Sam.
Then just distribute them you might get RHS

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Loser66
  • Loser66
@lesliesaree can you get it? I cannot. my answer is not the same your right hand side although I apply .Sam 's instruction
Loser66
  • Loser66
\[32(1-cos^2)(1-cos^2)(cos^2)= 32(1-cos^2)^2(cos^2)\] \[= 32(1-2cos^2 +cos^4)(cos^2)= 32(cos^2-2cos^4+cos^6)\] \[= 32cos^2 -64cos^4+32cos^6\] it's not the same your RHS
anonymous
  • anonymous
No I am not getting it because it needs to be reduced from the powers rather then ending with huge exponents but I keep getting stuck
Loser66
  • Loser66
your RHS is \[cos^2 x ~or~ cos (2x)\]
Loser66
  • Loser66
something else, just example
Loser66
  • Loser66
why didn't you ask him when you didn't get?
.Sam.
  • .Sam.
If RHS is \[2-\cos(2x)-2\cos(4x)+\cos(6x)\] Prove RHS = LHS is easier and Then you'll have to use some of these \[\Rightarrow \cos(2x)=1-2\sin^2(x) \\ \\ \Rightarrow \cos(2x)=2\cos^2(x)-1\] And \[\cos(A+B)=\cos(A)\cos(B)-\sin(A)\sin(B)\]
Loser66
  • Loser66
I am ok with this stuff, thanks Sam
anonymous
  • anonymous
Sorry :( but I was going to try that but I don't know how to reduce the ones that are bigger then 2 :o

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