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How would I solve the following double angle identity.

$$\cos^4x=\frac{3}{8}+\frac{1}{2}\cos(2x)+\frac{1}{8}\cos(4x)$$

So far my work is

$$\frac{3}{8}+\frac{2\cos^x-1}{2}+\frac{1}{8}(2\cos^2x-1)$$

But how would I proceed.

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Is "indentity" the new, cool way to spell that word these days? –  J. M. Jul 27 '12 at 15:52
    
I thought spell check correct it....... I think its spelled correctly. –  Fernando Martinez Jul 27 '12 at 15:53
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Off course which a spell chequer you can knot have any Miss spellings. :-) (BTW, my spell checker tells me "indentity" is wrong). –  celtschk Jul 27 '12 at 15:56
    
how is it spelled? –  Fernando Martinez Jul 27 '12 at 15:57
    
Use latex. Is it $\cos^4 x = \frac{3}{8} + \frac{1}{2}\cos 2x + \frac{1}{8}\cos 4x$? Where did you get $\cos^x$ in your work? –  Karolis Juodelė Jul 27 '12 at 15:58
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2 Answers 2

up vote 2 down vote accepted

Notice that \begin{eqnarray} \cos(2x)&=& \cos^2 x - \sin^2 x \\ &=& 2 \cos^2 x - 1.\\ \end{eqnarray} Then \begin{equation} \cos^2 x = \dfrac{1}{2}(1+\cos(2x)). \end{equation} Hence, \begin{eqnarray} \cos^4 x &=& (\cos^2 x)^2\\ &=& \left[\dfrac{1}{2}(1 + \cos(2x))\right]^2\\ &=& \dfrac{1}{4}(1 +2 \cos(2x)+ \cos^2(2x))\\ &=& \dfrac{1}{4} +\dfrac{1}{2} \cos(2x) + \dfrac{1}{4}\dfrac{1}{2}(1+\cos(4x))\\ &=& 3/8 + 1/2 \cos(2x) +1/8 \cos(4x) \end{eqnarray}

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I never knew the problem could be done in such a way. –  Fernando Martinez Jul 27 '12 at 16:06
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\begin{align*} \cos^4(x) &= \left(\frac{e^{ix}+e^{-ix}}{2}\right)^4\\ &= \frac{e^{4xi} + 4e^{2xi} + 6 + 4e^{-2xi} + e^{-4xi}}{16}\\ &= \frac{3}{8} + \frac{1}{2} \frac{e^{2xi} + e^{-2xi}}{2} + \frac{1}{8} \frac{e^{4xi}+e^{-4xi}}{2}\\ &= \frac{3}{8} + \frac{1}{2} \cos(2x) + \frac{1}{8} \cos(4x) \end{align*}

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Complex methods may not be what the OP had in mind, but they are a marvelous tool (+1). –  robjohn Aug 3 '12 at 20:03
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