lowering powers

May 2010
5
0
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
 

Soroban

MHF Hall of Honor
May 2006
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Lexington, MA (USA)
Hello, reino17!

Your algebra is off . . .


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

 
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May 2010
5
0
ahh ok i see where i screwed up. thanks dude