Problem :
Given that $\sin^3 x \sin 3x = \sum^n_{m=0}C_m \cos mx, C_n \neq 0 $ is an identity. Find the value of n.
I tried : $\sin3x = 3\sin x - 4\sin^3 x$ but unable to reach to any point.... Please suggest further ....Thanks..
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hint: you get $sin^3xsin3x=3sin^4x-4sin^6x$ but we also know that $sin^2(x)+cos^2(x)=1$ and that $cos(3x)=4cos^3x-3cosx$. You should combine between the identities. – Jun 03 '13 at 17:48
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Using the identity $$ \sin^3(x)=\frac{3\sin(x)-\sin(3x)}{4} $$ and the identity $$ \sin(ax)\sin(bx)=\frac12\big(\cos((a-b)x)-\cos((a+b)x)\big) $$ we get $$ \begin{align} \sin^3(x)\sin(3x) &=\frac{3\sin(x)-\sin(3x)}{4}\sin(3x)\\ &=\frac38\big(\cos(2x)-\cos(4x)\big)-\frac18\big(\cos(0x)-\cos(6x)\big)\\ &=\frac18\big(\cos(6x)-3\cos(4x)+3\cos(2x)-\cos(0x)\big) \end{align} $$ Thus, $n=6$.
robjohn
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2$$\sin(ax)\sin(bx)=\frac12\big(\cos((a-b)x)-\cos((a+b)x)\big)$$ has a cleaner form $\sin A\sin B=\frac12\cdot \big(\cos(A-B)-\cos(A+B)\big)$ – lab bhattacharjee Jun 03 '13 at 18:14
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1@labbhattacharjee: simpler in form, but essentially the same. However, I customized it to fit the problem at hand. It didn't seem an excessive crime :-) – robjohn Jun 03 '13 at 18:19
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HINT:
As you have noticed, $4\sin^3x=3\sin x-\sin3x,$
$$\text{So, }\sin^3x\sin3x=\frac{(3\sin x-\sin3x)\sin3x}4=\frac{3(2\sin x\sin3x)-2\sin^23x}8$$
As $2\sin A\sin B=\cos(A-B)-\cos(A+B),$
$\implies 2\sin x\sin3x=\cos(3x-x)-\cos(3x+x)=\cos2x-\cos4x$
and as $\cos2y=1-2\sin^2y\implies 2\sin^2y=1-\cos2y,$
$\implies 2\sin^23x=1-\cos6x$
lab bhattacharjee
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A related problem. Note that, the function $ \sin^3 x \sin 3x $ is an even function, so its Fourier series will have expansion in terms of the functions $\cos mx$ over the interval $[-\pi,\pi]$, where
$$ C_m = \frac{1}{\pi}\int_{-\pi}^\pi \sin^3 x \sin 3x \cos(mx)\, dx=\frac{2}{\pi}\int_{0}^\pi \sin^3 x \sin 3x \cos(mx)\, dx, \quad m \ge 0.$$
Now, just work out the integral and you will be able to finish the problem. See here for the value of the integral and you will see that the Fourier series has finite number of terms.
Note: This solution is based on the Fourier series knowledge. I just paid attention to the tag.
Mhenni Benghorbal
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1(+1) This is the general way to work these kinds of problems, just a bit advanced for the apparent level of the question. For instance, it's not certain that integrals are in scope. – robjohn Jun 04 '13 at 02:01
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@robjohn: Yes, you are right. You got the point. Thanks for the comment. – Mhenni Benghorbal Jun 04 '13 at 03:00
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$n=6$.
Since $\sin^3x \sin 3 x$ is a degree $6$ polynomial of $\cos x$, and $\cos nx$ is degree $n$ polynomial of $\cos x$, $n$ must be $6$.
Ma Ming
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Some more explanation might be nice, since the same reasoning was just used to justify $n=4$. – robjohn Jun 03 '13 at 18:22
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Why is $\sin^3(x)\sin(3x)$ a polynomial of $\cos(x)$? Given that, why is it degree $6$? Why is $\cos(nx)$ a degree $n$ polynomial of $\cos(x)$? These things are probably not obvious to a trigonometry student (I am making this assumption because of the question and the tags). – robjohn Jun 03 '13 at 19:03
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@robjohn: You are right. I think this my solution is a little bit advance for the OP. I just paid attention to the tag. However it is a correct solution. – Mhenni Benghorbal Jun 03 '13 at 19:17
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@MhenniBenghorbal: Your answer is a bit advanced for the level of the question, but I think you've intercepted someone else's message. – robjohn Jun 03 '13 at 22:51