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Prove that: $$\mathrm{cosec}\frac {\alpha}{8}+\mathrm{cosec}\frac {\alpha}{4}+\mathrm{cosec}\frac {\alpha}{2}=\cot \frac {\alpha}{16} - \cot \frac {\alpha}{2}$$.

My Attempt:

$$\text{L.H.S}=\mathrm{cosec}\frac {\alpha}{8}+\mathrm{cosec}\frac {\alpha}{4} + \mathrm{cosec}\frac {\alpha}{2}$$ $$=\frac {1}{\sin(\alpha/8)} + \frac {1}{2\sin(\alpha/8)\cdot \cos(\alpha/8)} + \frac {1}{\sin (\alpha/2)}$$. $$=\frac {2\cos (\alpha/8) +1}{2\sin(\alpha/8)\cdot\cos(\alpha/8)} + \frac {1}{\sin(\alpha/2)}$$.

How should I move on further? Please help.

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    $\begingroup$ Hint: $\csc x + \cot x = \frac{1+\cos x}{\sin x} = \cot 2x$. $\endgroup$
    – rogerl
    Nov 21, 2016 at 14:35
  • $\begingroup$ @rogerl, How do I use this? $\endgroup$
    – pi-π
    Nov 21, 2016 at 14:37

2 Answers 2

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Alternatively, one may use $$ \csc \frac {\alpha}{2}= \cot \alpha - \cot \frac {\alpha}{2} $$ giving $$ \csc \frac{\alpha}{2^{n}}= \cot \frac {\alpha}{2^{n+1}} - \cot \frac {\alpha}{2^{n}},\qquad n=1,2,3,\cdots, $$ then summing from $n=1$ to $n=3$ and using a telescoping sum gives $$ \csc \frac {\alpha}{8}+\csc \frac {\alpha}{4}+\csc \frac {\alpha}{2}=\cot \frac {\alpha}{16} - \cot \frac {\alpha}{2}. $$

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  • $\begingroup$ Do you have any other simpler method? I am not familiar with this process as it is not related to my grade. $\endgroup$
    – pi-π
    Nov 21, 2016 at 14:58
  • $\begingroup$ +1, Simpler than this one? $\endgroup$
    – Alex Silva
    Nov 21, 2016 at 15:07
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Note that in general $$\csc x + \cot x = \frac{1+\cos x}{\sin x} = \cot 2x.$$ Now, moving $\cot\frac{\alpha}{2}$ to the LHS and applying this identity multiple times gives \begin{align} \csc\frac {\alpha}{8} + \csc\frac {\alpha}{4} + \csc\frac {\alpha}{2} + \cot \frac {\alpha}{2} &= \csc \frac{\alpha}{8} + \csc\frac{\alpha}{4} + \cot\frac{\alpha}{4} \\ &= \csc \frac{\alpha}{8} + \cot\frac{\alpha}{8} \\ &= \cot \frac{\alpha}{16}. \end{align}

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