Question

Prove the following equation:
Will post question below.

Answer 1

\[\large \sin^2 \theta \cos^2 \theta = \frac{ 1 }{ 8 }(1-\cos 4\theta)\]

Answer 2

@amistre64 can you please help me??

Answer 3

how to prove? work left to right?

Answer 4

some things to keep in mind:

\[cos(2x)=cos^2(x)-sin^2(x)\]
\[cos(2x)=1-2sin^2(x)\]
\[cos(2x)=2cos^2(x)-1\]

Answer 5

\[2sin(x)cos(x)=sin(2x)\]

Answer 6

\[cos(4x)=1-2sin^2(2x)\]
\[2sin^2(2x)=1-cos(4x)\]
\[sin^2(2x)=\frac{1}{2}(1-cos(4x))\]
\[[2sin(x)cos(x)]^2=\frac{1}{2}(1-cos(4x))\]
\[4sin^2(x)cos^2(x)=\frac{1}{2}(1-cos(4x))\]
\[sin^2(x)cos^2(x)=\frac{1}{8}(1-cos(4x))\]
if you have to back into it that should give you some direction

Answer 7

thank you very much!

Answer 8

youre welcome