Question
$\cos\frac{2\pi}{15}\cos\frac{4\pi}{15}\cos\frac{8\pi}{15}\cos\frac{16\pi}{15}=\frac{1}{16}$
✓
Answer
True.
Solution:
$\text{L.H.S.}\cos\frac{2\pi}{15}\cdot\cos\frac{4\pi}{15}\cdot\cos\frac{8\pi}{15}\cdot\cos\frac{16\pi}{15}$
$=\cos24^\circ.\cos48^\circ.\cos96^\circ.\cos192^\circ$
$=\frac{1}{16\sin24^\circ}[2\sin24^\circ\cos24^\circ)(2\cos48^\circ)(2\cos96^\circ)(2\cos192^\circ)]$
$=\frac{1}{16\sin24^\circ}[\sin48^\circ\cdot2\cos48^\circ(2\cos96^\circ)(2\cos192^\circ)]$
$=\frac{1}{16\sin24^\circ}[2\sin48^\circ\cos48^\circ(2\cos96^\circ)(2\cos192^\circ)]$
$\frac{1}{16\sin24^\circ}[\sin96^\circ(2\cos96^\circ)(2\cos192^\circ)]$
$=\frac{1}{16\sin24^\circ}[2\sin96^\circ\cdot\cos96^\circ(2\cos192^\circ)]$
$=\frac{1}{16\sin24^\circ}[\sin192^\circ\cdot(2\cos192^\circ)]$
$=\frac{1}{16\sin24^\circ}2\sin192^\circ\cos192^\circ$
$=\frac{1}{16\sin24^\circ}\sin384^\circ=\frac{1}{16\sin24^\circ}\sin(360^\circ+24^\circ)$
$=\frac{1}{16\sin24^\circ}\times\sin24^\circ[\because\sin(360^\circ+\theta)=\sin\theta]$
$=\frac{1}{16}\text{R.H.S.}$
Hence, the given statement is 'True'.
Solution:
$\text{L.H.S.}\cos\frac{2\pi}{15}\cdot\cos\frac{4\pi}{15}\cdot\cos\frac{8\pi}{15}\cdot\cos\frac{16\pi}{15}$
$=\cos24^\circ.\cos48^\circ.\cos96^\circ.\cos192^\circ$
$=\frac{1}{16\sin24^\circ}[2\sin24^\circ\cos24^\circ)(2\cos48^\circ)(2\cos96^\circ)(2\cos192^\circ)]$
$=\frac{1}{16\sin24^\circ}[\sin48^\circ\cdot2\cos48^\circ(2\cos96^\circ)(2\cos192^\circ)]$
$=\frac{1}{16\sin24^\circ}[2\sin48^\circ\cos48^\circ(2\cos96^\circ)(2\cos192^\circ)]$
$\frac{1}{16\sin24^\circ}[\sin96^\circ(2\cos96^\circ)(2\cos192^\circ)]$
$=\frac{1}{16\sin24^\circ}[2\sin96^\circ\cdot\cos96^\circ(2\cos192^\circ)]$
$=\frac{1}{16\sin24^\circ}[\sin192^\circ\cdot(2\cos192^\circ)]$
$=\frac{1}{16\sin24^\circ}2\sin192^\circ\cos192^\circ$
$=\frac{1}{16\sin24^\circ}\sin384^\circ=\frac{1}{16\sin24^\circ}\sin(360^\circ+24^\circ)$
$=\frac{1}{16\sin24^\circ}\times\sin24^\circ[\because\sin(360^\circ+\theta)=\sin\theta]$
$=\frac{1}{16}\text{R.H.S.}$
Hence, the given statement is 'True'.
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