# Pls help to prove cos^2(2θ)-cos^2(6θ)=sin4θ*sin8θ. Thanks

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### 1 Answer

`cos^2(2theta) - cos^2(6theta) = sin(4theta)*sin(8theta)`

We can use

`cos(A+B) = cos(A)cos(B) - sin(A)sin(B)` and

`cos(A-B) = cos(A)cos(B) + sin(A)sin(B)` if we subtact these two

`cos(A+B) - cos(A-B) = -2sin(A)sin(B)`

so if `A=8theta` and `B=4theta` we get

`cos(8theta+4theta) - cos(8theta-4theta) = -2sin(8theta)sin(4theta)` simplifying

`cos(12theta) - cos(4theta) = -2sin(4theta)sin(8theta)` or

`cos(4theta) - cos(12theta) = 2sin(4theta)sin(8theta)` or

`1/2(cos(4theta) - cos(12theta)) = sin(4theta)sin(8theta)` so we have

`cos^2(2theta) - cos^2(6theta) = sin(4theta)sin(8theta) = 1/2 (cos(4theta) - cos(12theta))` and now we use the double angle formula

`cos(2A)=2cos^2(A) - 1`

`cos^2(2theta)-cos^2(6theta) = 1/2(2cos^2(2theta) - 1 - (2cos^2(6theta)) `

`cos^2(2theta)-cos^2(6theta) = 1/2(2cos^2(2theta) - 1 - 2cos^2(6theta) + 1) `

`cos^2(2theta)-cos^2(6theta) = 1/2(2cos^2(2theta) - 2cos^2(6theta)) `

`cos^2(2theta)-cos^2(6theta) = cos^2(2theta) - cos^2(6theta)`

This was what we wanted to prove...