rewrite the expression in terms of the first power of cosine

sin^4x

so far i have

(sin^2x)^2

= (1-cos2x/2)^2

= (1-cos2x/2)(1-cos2x/2)

=1-cos^2.2x/4

so my question is, is it right? or is there more

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- May 16th 2010, 07:42 PMreino17lowering powers
rewrite the expression in terms of the first power of cosine

sin^4x

so far i have

(sin^2x)^2

= (1-cos2x/2)^2

= (1-cos2x/2)(1-cos2x/2)

=1-cos^2.2x/4

so my question is, is it right? or is there more - May 16th 2010, 08:32 PMSoroban
Hello, reino17!

Your algebra is off . . .

Quote:

Rewrite the expression in terms of the first power of cosine: .$\displaystyle \sin^4x$

$\displaystyle \sin^4\!x \;\;=\;\;\left(\sin^2\!x\right)^2 \;\;=\;\;\left(\frac{1-\cos2x}{2}\right)^2 \;\;=$ . $\displaystyle \frac{1-2\cos2x + \cos^2\!2x}{4}\;\;=\;\;\frac{1 - 2\cos2x + \frac{1 + \cos4x}{2}}{4} $

Multiply by $\displaystyle \tfrac{2}{2}:\quad \frac{2-4\cos2x + 1 + \cos4x}{8} \;\;=\;\;\tfrac{1}{8}(3 - 4\cos 2x + \cos4x) $

- May 16th 2010, 09:03 PMreino17
ahh ok i see where i screwed up. thanks dude