# Expressing $\sin^4x-\sin^6x$ in another way

I am slightly confused on how one would subtract $\sin^4x-\sin^6x$.

I know that $\sin^2x=(1/2)(1-\cos2x)$,

so $\sin^4x$ would logically be $[(1/2)(1-\cos2x)]^2=(1/4)(1-2\cos(2x)+\cos^2(2x)$

However the value of $\sin^6x$ eludes me. Would it be $(1-\cos2x)^3$? I did that and got $1-\cos2x-2\cos2x-2\cos^2(2x)+\cos^2(2x)-\cos^3(2x)$

I am not sure if that is correct.

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It's a bit ambiguous to discuss "how to subtract" trigonometric terms; as others have noted, it all depends on what kind of form you want to end up with. –  K. Hu Aug 8 '12 at 17:39
This maybe useful to you: bit.ly/PFTcPc –  Vitaliy Kaurov Aug 8 '12 at 22:44

I have noticed the other answers either don't answer your question (and get many upvotes none-the-less) or are much more complicated than necessary (even more complicated than you understand based on the comment by Raymond). So, here is the answer to your actual question.

You dropped the 1/2.

$$\sin^6 x = (\sin^2 x)^3 = \left(\frac{1 - \cos{2x}}{2}\right)^3 = \frac{1}{8} (1 - \cos{2x})^3$$

so you end up missing $1/8$ in the end. Also, you messed up the sign on one term. In general

$$(A + B)^3 = A^3 + 3A^2B + 3AB^2 + B^3$$

so here we have

\begin{align*} (1 - \cos(2x))^3 &= \bigg(1 + (-\cos(2x))\bigg)^3 \\ &= 1 + 3(-\cos(2x)) + 3(-\cos(2x))^2 + (-\cos(2x))^3 \\ &= 1 - 3\cos(2x) + 3\cos^2(2x) - \cos^3(2x). \end{align*}

\begin{align*} \sin^4x - \sin^6x &= \frac{1}{4}\bigg(1 - 2\cos(2x) + \cos^2(2x)\bigg) \\ &- \frac{1}{8}\bigg(1 - 3\cos(2x) + 3\cos^2(2x) - \cos^3(2x)\bigg) \\ &= \frac{1}{8}\bigg(1 - \cos(2x) - \cos^2(2x) + \cos^3(2x)\bigg) \end{align*}

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"Your final answer"? For someone who wrote in a comment above that isn't sure what the OP is asking you seem to be pretty sure about it...And the others got upvoted because they gave good answers to at least part of the confusing question the OP asked. If there's some doubt about what it is that's the OP's responsibility. –  DonAntonio Aug 8 '12 at 20:24
@DonAntonio I didn't say I wasn't sure what the OP is asking. I said "I admit it's slightly hard to tell." which means you need to take some time to read their question and see that they are trying to complete the calculation I showed, yet got stuck and needed help getting through it. If you quickly read the question, or just read the title, you will think the question is just asking for various ways to write the expression in different forms. But, the title just gives you the general area of the problem and you read the question to see what is really meant, which I did, and others did not. –  Graphth Aug 8 '12 at 20:51
And, I'm not saying their answers aren't interesting or helpful. They are. And, maybe they are what the OP is looking for. –  Graphth Aug 8 '12 at 21:18
Yes having scanned the answers even through I enjoyed looking at them I think this is the one that most suits what I was asking. –  Fernando Martinez Aug 8 '12 at 22:05
One question about the final answer is that would not -2cos(2x)-(-3cos(2x)=1cos(2x)? –  Fernando Martinez Aug 8 '12 at 22:22

There are various ways to transform $\sin^4 x-\sin^6 x$ to some equivalent form. For example, $\sin^4 x-\sin^6 x=\sin^4 x(1-\sin^2 x)=\sin^4 x\,\cos^2 x$.

If you are interested in expressing the difference as a sum of trigonometric functions of multiples of $x$, this will provide an efficient start. For then you can use $\sin^2 x=\frac{1-\cos 2x}{2}$ and $(\sin x\cos x)^2=\frac{(\sin 2x)^2}{4}$.

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Doesn't answer the question. Answers the title. I admit it's slightly hard to tell what the OP is asking. And, this could be instructive. –  Graphth Aug 8 '12 at 19:04

Rather than messing about with trig identities, it may be simpler to use complex exponentials. If $z = e^{ix} = \cos(x) + i \sin(x)$, so $\sin(x) = (z - z^{-1})/(2i)$ and $\cos(x) = (z + z^{-1})/2$, \eqalign{\sin^4 x - \sin^6 x &= \frac{(z - z^{-1})^4}{16} + \frac{(z - z^{-1})^6}{64}\cr &= \frac{z^4 - 4 z^2 + 6 - 4 z^{-2} + z^{-4}}{16} \\ &\qquad+\frac{z^6 - 6 z^4 + 15 z^2 - 20 + 15 z^{-2} - 6 z^{-4} + z^{-6}}{64}\cr &= \frac{z^6 + z^{-6}}{64} - \frac{z^4 + z^{-4}}{32} - \frac{z^2 + z^{-2}}{64} + \frac{1}{16}\cr &= \frac{\cos(6x)}{32} - \frac{\cos(4x)}{16} - \frac{\cos(2x)}{32} + \frac{1}{16}\cr}

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$\sin^4x-\sin^6x=\sin^4x(1-\sin^2x)=\sin^4x \cos^2x=\sin^2x(\sin^2x \cos^2x)$ $$=\frac{\sin^2 x\,\sin^2 2x}{4}=\left(\frac{\sin x \sin 2x}{2}\right)^2=\left(\frac{\cos x-\cos3x}{4}\right)^2$$

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