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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.

flodyninja
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Fernando Martinez
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4 Answers4

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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 $(\sin x\cos x)^2=\frac{(\sin 2x)^2}{4}$.

André Nicolas
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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. – GeoffDS Aug 08 '12 at 19:04
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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 $\cos(x) = (z + z^{-1})/2$, $$\eqalign{\sin^4 x - \sin^6 x &= \frac{(z - z^{-1})^4}{16} + \frac{(z - z^{-1})^6}{64}\cr &= \frac{z^4 - 4 z^2 + 6 - 4 z^{-2} + z^{-4}}{16} \\ &\qquad+\frac{z^6 - 6 z^4 + 15 z^2 - 20 + 15 z^{-2} - 6 z^{-4} + z^{-6}}{64}\cr &= \frac{z^6 + z^{-6}}{64} - \frac{z^4 + z^{-4}}{32} - \frac{z^2 + z^{-2}}{64} + \frac{1}{16}\cr &= \frac{\cos(6x)}{32} - \frac{\cos(4x)}{16} - \frac{\cos(2x)}{32} + \frac{1}{16}\cr}$$

hmakholm left over Monica
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Robert Israel
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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*}

Therefore, your final answer is

\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*}

GeoffDS
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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 08 '12 at 20:24
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    @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. – GeoffDS Aug 08 '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. – GeoffDS Aug 08 '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 08 '12 at 22:05
  • One question about the final answer is that would not -2cos(2x)-(-3cos(2x)=1cos(2x)? – Fernando Martinez Aug 08 '12 at 22:22
  • @FernandoMartinez Don't forget the fractions out in front. It's (1/4) * -2 + (-1/8) * (-3) = -1/2 + 3/8 = -1/8. Is that the part you're asking about? The coefficient of the $\cos(2x)$ term? – GeoffDS Aug 08 '12 at 22:47
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$\sin^4x-\sin^6x=\sin^4x(1-\sin^2x)=\sin^4x \cos^2x=\sin^2x(\sin^2x \cos^2x)$ $$=\frac{\sin^2 x\,\sin^2 2x}{4}=\left(\frac{\sin x \sin 2x}{2}\right)^2=\left(\frac{\cos x-\cos3x}{4}\right)^2$$

Vincenzo Tibullo
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Aang
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