Question

please help me, i dont know how to start this equation

sinx^2xcosx^2x=(1-cos4x)/8

Answer 1

you just need to prove that they are identities, by expanding the sinx^2xcosx^2x

Answer 2

\[\sin ^{2}x \cos ^{2}x=(1-\cos4x)\div8\]

Answer 3

Try using \[\cos2x=2\cos^2x - 1\]

Answer 4

That can work when you start on the right hand side. Down the road you will also need:\[\cos 2x=1-2\sin ^{2}x\] To get started you will need to use substitution involving 2x with mboorstin's identity that will look like:\[\cos 4x=2\cos ^{2}2x-1\]This route involves recognizing a difference of squares and ultimately using the same original identities in reverse to get what you are looking for.

I preferred to start on the left hand side by substituting:\[\sin ^{2}x=\frac{1}{2}(1-\cos 2x)\]and\[\cos ^{2}x=\frac{1}{2}(1+\cos2x)\]Multiply it out (don't forget FOIL). Use the cosine squared formula again to get up to 4x. Then simplify and factor the fractions from it to make it look like the right hand side.

Answer 5

Oh yeah... and watch your minus signs when distributing.