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Question: Find the sum of the series: $$\cos^3 \alpha +\cos^3 {3\alpha} + \cos^3 {5\alpha}+....+\cos^3 {(2n-1)\alpha}$$

The book from which this question was taken says that the answer is $\frac{3\sin{n\alpha}\cos{n\alpha}}{4\sin\alpha}+\frac{\sin{3n\alpha}\cos{3n\alpha}}{4\sin{3\alpha}}$.

My attempt to solve this question:
$$\text{Let S be the trigonometric series,}$$ $$\cos {3\theta} = 4\cos^3\theta-3\cos\theta \implies 4\cos^3 \theta=\cos{3\theta}+3\cos\theta$$ Applying formula on $\cos^3\theta$,.. $$4S = 3\cos\alpha + \cos3\alpha +3\cos3\alpha + \cos9\alpha+...+3\cos{(2n-1)\alpha}+\cos{(6n-3)\alpha}$$ $$4S= 3(\cos \alpha + \cos 3\alpha+\cos5\alpha+...)+(\cos3\alpha + \cos9\alpha+..)$$ Applying the summation of cosine formula, $$4S= 3\frac{\sin{n\alpha}}{\sin\alpha}\cdot\cos{(\alpha+(n-1)\alpha)}+?$$

So my problem is I don't know how to apply the formula for the second series (denoted by '?') for a fixed number of terms $n$ or should I treat it as a general and separate series.

To clarify my doubt again, what I am asking is that would the $(2n-1)$ affect the formula for the second series (denoted by '?').

My working here looks most probably correct but if there is any mistake please correct it.

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  • $\begingroup$ math.stackexchange.com/questions/17966/… $\endgroup$ – lab bhattacharjee May 8 '19 at 3:57
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    $\begingroup$ Trigonometric sum $(\cos \alpha + \cos 3\alpha+\cos5\alpha+...)$ is the real part of $e^{i\alpha}+e^{3i\alpha}+e^{5i\alpha}+...=t(1+t^2+t^4+...)$, a geometric sum with $t=e^{i\alpha}$. Similar operation for the other trigonometric sum. $\endgroup$ – Jean Marie May 8 '19 at 4:59
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    $\begingroup$ math.stackexchange.com/questions/17966/… $\endgroup$ – lab bhattacharjee May 25 '19 at 6:25
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    $\begingroup$ This is the best question in the whole world $\endgroup$ – rash May 25 '19 at 6:28
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    $\begingroup$ somebody should answer it, fraud??? $\endgroup$ – rash May 25 '19 at 6:28
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If $\sin3\alpha=0$ it's smooth.

But for $\sin3\alpha\neq0$ by the telescopic sum we obtain: $$\sum_{k=1}^n\cos^3(2k-1)\alpha=\frac{1}{4}\sum_{k=1}^n(\cos3(2k-1)\alpha+3\cos(2k-1)\alpha)=$$ $$=\frac{\sum\limits_{k=1}^n2\sin3\alpha\cos(6k-3)\alpha}{8\sin3\alpha}+\frac{3\sum\limits_{k=1}^n2\sin\alpha\cos(2k-1)\alpha}{8\sin\alpha}=$$ $$=\frac{\sum\limits_{k=1}^n(\sin6k\alpha-\sin(6k-6)\alpha)}{8\sin3\alpha}+\frac{3\sum\limits_{k=1}^n(\sin2k\alpha-\sin(2k-2)\alpha)}{8\sin\alpha}=$$ $$=\frac{\sin6n\alpha}{8\sin3\alpha}+\frac{3\sin2n\alpha}{8\sin\alpha}.$$

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Hint: Prove by induction that $$\sum_{i=1}^n\cos((2i-1)\alpha)^3=\frac{1}{8} \left(-\csc (3) \cos \left(\frac{1}{2} (-12 n-\pi )\right)-3 \csc (1) \cos \left(\frac{1}{2} (-4 n-\pi )\right)\right)$$

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Just take $\beta = 3\alpha$ in the 2nd part, and apply the same formula as in the first part.

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