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

`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...