Use the power-reducing formulas as many times as possible to rewrite the expression in terms of the first power of the cosine. sin^2 3x cos^2 3x
@3mar Do you know how to do this the easy way?
First us that \[ \sin(a) \cos(a)= \frac 1 2 \sin(2a) \]
This gives \[ \sin^2 3x \cos^2 3x=\frac 1 4 \sin^2(6x) \]
14sin2(6x)=sin2(3x)cos2(3x)
Use now the formula of reduction of power and you are done
hold on just a second
Can you start from this step?
\[\sin ^23x \cos ^23x=\frac{ 4\sin ^23x \cos ^23x }{ 4 }=\frac{ \left( 2\sin 3x \cos 3x \right)^2 }{ 4 }\] \[=\frac{ \sin ^26x }{ 4 }=\frac{ 2\sin ^26x }{ 8 }=\frac{ 1-\cos 12x }{ 8 }\] [sin 2y=2 sin y cos y] \[1-\cos 2y=2\sin ^2y\]
So how can I get the final answer from there?
Now use \[ \sin ^2(u)=\frac{1}{2} (1-\cos (2 u)) \]
@eliesaab Which one do I plug in?
here u=6x as eliesaab told and y=6x as i have told.
\[\frac{ 1 }{ 2 }\] (1−cos(2)(6x) @sshayer @eliesaab
4*2=8
Yes! @sshayer
@satellite73
@TheMathAsker Do you know how to get the final answer?
i don't know what you want in the answer.
I think I figured it out, thank you!
\[\frac{ 1 }{ 8 }\left( 1-\cos 12x \right)\]
ok.yw
Sorry, I was not here! But I think @sshayer did it very well
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