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.
Answer
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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