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The problem reads as follows.

Prove this identity: $$\sin^4x = \frac{1}{8}(3 - 4\cos2x + \cos4x)$$

I started with the right side and used double angles identities for $\cos2x$ and a sum and then then double angle identity for the $\cos4x...$ It all got messy and I hit a dead end. He doesn't give any hints and I'm pretty lost.

I'm thinking that I will eventually need a product-sum identity to get the $\dfrac{1}{8}....$ But I'm just confused how to get there...

Thanks in advance to anyone who can help!

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  • $\begingroup$ Parentheses, please. It is not clear that the last term is in the numerator, not the denominator. Here, from context, it must be. But we see many $1/2x$s, some of which are $x/2$ and some $1/(2x)$ Better yet, use $\LaTeX$ and stack the fractions. A tutorial is here Anupam did it for you-you can review the edit history to see how. $\endgroup$ – Ross Millikan Apr 21 '14 at 3:05
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Or better yet, use Euler's formula!

$$\sin^4 x=\left(\frac{e^{ix}-e^{-ix}}{2i}\right)^4={1\over 16}\left(e^{4ix}-4e^{2ix}+6-4e^{-2ix}+e^{-4ix}\right)={1\over 8}\left(\frac{e^{4ix}+e^{-4ix}}{2}-4\frac{e^{2ix}+e^{-2ix}}{2}+3\right)={1\over 8}\left(\cos{4x}-4\cos{2x}+3\right)$$ as desired.

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\begin{align} \sin^4 x & = \left(\frac{1-\cos 2x}{2}\right)^2 \\ & = \frac{1-2\cos 2x + \cos^2 2x}{4} \\ & = \frac{1}{4}\left(1-2\cos 2x+\frac{1+\cos 4x}{2}\right) \\ & = \frac{1}{8}\left(3-4\cos 2x +\cos 4x\right) \end{align}

Where identities $\sin^2 t=\frac{1-\cos (2t)}{2}$ and $\cos^2 t=\frac{1+\cos (2t)}{2}$ were used.

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Do you know power-reduction formulas?

$$\sin^2(x) = \frac{1-\cos(2x)}{2}$$ $$\cos^2(x) = \frac{1 + \cos(2x)}{2}$$

From here, it is fairly simple once we recognize that $\sin^4(x) = (\sin^2(x))^2$.

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$\sin^4x = (1 - \cos^2x)^2 = \left(1 - \dfrac{1 + \cos2x}{2}\right)^2 = \dfrac{(1 - \cos2x)^2}{4} = \dfrac{1 - 2\cos2x + \cos^22x}{4} = \dfrac{1}{4} - \dfrac{\cos2x}{2} + \dfrac{\cos^22x}{4} = \dfrac{1}{4} - \dfrac{\cos2x}{2} + \dfrac{\dfrac{1 + \cos4x}{2}}{4} = \dfrac{1}{4} - \dfrac{\cos2x}{2} + \dfrac{1 + \cos4x}{8} = \dfrac{3 - 4\cos2x + \cos4x}{8}$.

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    $\begingroup$ Hey, you're finally learning latex! You should put a \ in front of sin and cos to give them non-italicized letters, e.g. $\sin x$ vs. $sin x$ $\endgroup$ – MCT Apr 21 '14 at 3:21
  • $\begingroup$ @MichaelT: cheers. It takes longer to figure it out. I looked at "show math as" to see what symbols other users used to type their posts. $\endgroup$ – DeepSea Apr 21 '14 at 5:12
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Everyone starts from LHS, I choose to start from RHS. \begin{align} \dfrac{1}{8}(3 - 4\cos2x + \cos4x)&=\dfrac{1}{8}(3 - 4\cos2x + 2\cos^2 2x-1)\\ &=\dfrac{1}{8}(2 - 4\cos2x + 2\cos^22x)\\ &=\dfrac{2}{8}(1 - 2\cos2x + \cos^22x)\\ &=\dfrac{1}{4}(1-\cos2x)^2\\ &=\dfrac{1}{4}(1-2\cos^2x+1)^2\\ &=\dfrac{1}{4}2^2(1-\cos^2x)^2\\ &=\sin^4x. \end{align}

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