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Folks, how can $\sin^4(x)$ be simplified to a trig function with power of 1? I tried: $(\sin^2(x))^2 = ((1-\cos2x)/2)^2$ but still getting $\cos x$ to the power of $2$. Wolframalpha only shows the answer which i have no idea how they got it. Could somebody help ?

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    $\begingroup$ This is the correct first step. Now you have to do the same for $\cos^2(2x)$. $\endgroup$ – Marcus M Jul 22 '15 at 16:03
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Using $\cos2y=1-2\sin^2y,$

$$(2\sin^2x)^2=(1-\cos2x)^2=1-2\cos2x+\cos^22x$$

Use $\cos2y=2\cos^2y-1$ for $\cos^22x$


Alternatively using Euler's formula $2i\sin y=e^{iy}-e^{-iy};2\cos y=e^{iy}+e^{-iy}$

$$\sin^4x=\left(\dfrac{e^{ix}-e^{-ix}}{2i}\right)^4$$

$$\displaystyle=\dfrac{e^{i4x}+e^{-i4x}-\binom41(e^{i2x}+e^{-i2x})+\binom42}{16}=?$$

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  • $\begingroup$ Thanks for the response, but the problem has be simplified so that any trig function would have only a power of 1 while we're getting cos^2 (2x) $\endgroup$ – Meaghan Jul 22 '15 at 16:18
  • $\begingroup$ @Meaghan, Don't you notice : Use $\cos2y=2\cos^2y-1$ for $\cos^22x$ $\endgroup$ – lab bhattacharjee Jul 22 '15 at 16:19
  • $\begingroup$ But how can i use what you have mentioned ? $\endgroup$ – Meaghan Jul 22 '15 at 16:30
  • $\begingroup$ @Meaghan, If $\cos^2y=\dfrac{1+\cos2y}2,\cos^22x=?$ $\endgroup$ – lab bhattacharjee Jul 22 '15 at 16:31
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$\sin^4(x) = \left(\sin^4 x\right)^\color{red}{1}$. Jokes apart, $$\sin^4(x) = \left(1-\cos^2(x)\right)^2 = \left(\frac{1-\cos(2x)}{2}\right)^2=\frac{1}{4}-\frac{\cos(2x)}{2}+\frac{\cos^2(2x)}{4} $$ hence: $$\sin^4(x) = \frac{3}{8}-\frac{\cos(2x)}{2}+\frac{\cos(4x)}{8}=\frac{3-4\cos(2x)+\cos(4x)}{8}.$$

I have just applied the Pythagorean theorem ($\sin^2 z+\cos^2 z=1$) and twice the cosine duplication formula ($\cos(2z)=2\cos^2 z-1$, giving $\cos^2(z)=\frac{1+\cos(2z)}{2}$).

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  • $\begingroup$ Can you explain how you got the answer? $\endgroup$ – Meaghan Jul 22 '15 at 16:27
  • $\begingroup$ @Meaghan: done. $\endgroup$ – Jack D'Aurizio Jul 22 '15 at 16:35
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$$\sin^4 x = (\sin^2x)^2$$

By using identity $\sin^2 x = 1- \cos^2 x$, we can change $\sin^4 x$ to:

$$\sin^4 x = (1-\cos^2 x)^2$$

$\cos^2 x$ can be changed by using identity $\cos 2x= 2\cos^2 x-1$, then $\cos^2 x = \frac{1+\cos 2x}{2}$

So, $\sin^4 x = (1-\frac12-\frac12\cos 2x)^2$

$$=(\frac12-\frac12\cos 2x)^2$$ $$= \frac14(1-\cos 2x)^2$$ $$= \frac14(1-2\cos2x+\cos^2 2x)$$

Also change $\cos^2 2x$ by using identity $\cos 4x= \cos 2(2x)= 2\cos^2 2x-1$.

So $\cos^2 2x= \frac12(\cos 4x+1)$

$$= \frac14(1-2\cos2x +(\cos 4x+1)/2)$$ $$= \frac14(1-2\cos2x +\frac12\cos4x +\frac12)$$ $$= \frac14(3/2 -2\cos2x +\frac12\cos 4x)$$

Finally the answer should be:

$$\sin^4x= \frac38 - \frac12\cos(2x) + \frac18\cos(4x)$$

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The other answers have used a trigonometrical/algebraic approach to the problem, this answer showcases a different method that exploits the relationship between $z^n \pm z^{-n}$ for complex numbers $z$.

We know that $$z^n - z^{-n} = 2i\sin n\theta \implies z - z^{-1} = 2i\sin \theta \\ z^n +z^{-n} = 2\cos n\theta \implies z + z^{-1} = 2\cos \theta$$ from a basic application of De Moivre's theorem.

So, it follows immediately that $$\begin{align}(2i \sin \theta)^4 &= (z-z^{-1})^4 \\ &=z^4 + \frac{1}{z^4} - 4\left(z^2 + \frac{1}{z^2}\right) + 6 \\ &= 2\cos 4\theta - 8\cos 2\theta + 6\end{align}$$

So we have, finally $$16 \sin^4 \theta = 2\cos 4\theta - 8\cos 2\theta + 6$$

So dividing through by $16$ yields $$\bbox[border: solid 1px blue, 10px]{\sin^4 \theta = \frac{1}{8} \cos 4\theta - \frac{1}{2}\cos 2\theta + \frac{3}{8}}$$

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