Question

Prove: 32(sin^4x)(cos^2x)=2-cos2x-2cos4x+cos6x

Please help :(

Answer 1

The RHS has all the cosine function, you might wanna change the LHS sine function to cosine

Answer 2

\[32(\sin^4x)(\cos^2x)\]
\[\Rightarrow32(\sin^2(x)\sin^2(x))(\cos^2x)\]
Using
\[\sin^2x=1-\cos^2x\]
\[\Rightarrow32((1-\cos^2x)(1-\cos^2x))(\cos^2x)\]

Answer 3

Then just distribute them you might get RHS

Answer 4

@lesliesaree can you get it? I cannot. my answer is not the same your right hand side although I apply .Sam 's instruction

Answer 5

\[32(1-cos^2)(1-cos^2)(cos^2)= 32(1-cos^2)^2(cos^2)\]
\[= 32(1-2cos^2 +cos^4)(cos^2)= 32(cos^2-2cos^4+cos^6)\]
\[= 32cos^2 -64cos^4+32cos^6\] it's not the same your RHS

Answer 6

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

Answer 7

your RHS is \[cos^2 x ~or~ cos (2x)\]

Answer 8

something else, just example

Answer 9

why didn't you ask him when you didn't get?

Answer 10

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
\[\cos(A+B)=\cos(A)\cos(B)-\sin(A)\sin(B)\]

Answer 11

I am ok with this stuff, thanks Sam

Answer 12

Sorry :( but I was going to try that but I don't know how to reduce the ones that are bigger then 2 :o