Question

if 15x=π,prove that cos x cos 2x cos 3x cos 4x cos 5x cos6x cos7x=1/2^7

Answer 1

$\textbf{Given:}\;x=\frac{\pi}{15}$

$\textbf{To prove:}$

$cosx\;cos2x\;cos3x\;cos4x\;cos5x\;cos6x\;cos7x=\displaystyle\frac{1}{2^7}$

$\textbf{That is, to prove}$

$cos12^{\circ}\;cos24^{\circ}\;cos36^{\circ}\;cos48^{\circ}\;cos60^{\circ}\;cos72^{\circ}\;cos84^{\circ}=\displaystyle\frac{1}{2^7}$

$\text{Consider,}$

$cos12^{\circ}\;cos24^{\circ}\;cos36^{\circ}\;cos48^{\circ}\;cos60^{\circ}\;cos72^{\circ}\;cos84^{\circ}$

$=cos12^{\circ}\;cos24^{\circ}\;cos36^{\circ}\;cos48^{\circ}\displaystyle(\frac{1}{2})cos72^{\circ}\;cos84^{\circ}$

$=\displaystyle(\frac{1}{2})[cos48^{\circ}\;cos12^{\circ}\;cos72^{\circ}][\;cos36^{\circ}\;cos24^{\circ}\;cos84^{\circ}]$

$\text{We know that,}$

$\boxed{\bf\,cos(60-A)\;cosA\;cos(60+A)=\frac{1}{4}cos\,3A}}$

$=\displaystyle(\frac{1}{2})[\frac{1}{4}\;cos3(12^{\circ})][\frac{1}{4}\;cos3(24^{\circ})]$

$=\displaystyle(\frac{1}{2})[\frac{1}{4}\;cos36^{\circ}][\frac{1}{4}\;cos72^{\circ}]$

$=\displaystyle(\frac{1}{32})[cos36^{\circ}][cos72^{\circ}]$

$=\displaystyle(\frac{1}{32})[\frac{\sqrt{5}+1}{4}][\frac{\sqrt{5}-1}{4}]$

$=\displaystyle(\frac{1}{32})[\frac{(\sqrt{5})^2-1^2}{16}]$

$=\displaystyle(\frac{1}{32})[\frac{5-1}{16}]$

$=\displaystyle(\frac{1}{32})[\frac{4}{16}]$

$=\displaystyle(\frac{1}{32})[\frac{1}{4}]$

$=\displaystyle(\frac{1}{2^5})[\frac{1}{2^2}]$

$=\bf\displaystyle\frac{1}{2^7}$

$\therefore\bf\,cos12^{\circ}\;cos24^{\circ}\;cos36^{\circ}\;cos48^{\circ}\;cos60^{\circ}\;cos72^{\circ}\;cos84^{\circ}=\displaystyle\frac{1}{2^7}$

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Answer 2

Answer:

if 15x=π,prove that cos x cos 2x cos 3x cos 4x cos 5x cos6x cos7x=1/2^7