Question

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$prove \: that \: \\ 16 \cos \frac{2\pi}{15} \cos\frac{4\pi}{15} \cos \frac{8\pi}{15} \cos \frac{14\pi}{15} = 1$




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Answer 1

Step-by-step explanation:

We have,

$\tt{16 \: cos \bigg( \dfrac{2 \pi}{15} \bigg) \:cos \bigg( \dfrac{4 \pi}{15} \bigg) \:cos \bigg( \dfrac{8 \pi}{15} \bigg) \:cos \bigg( \dfrac{14 \pi}{15} \bigg) }$

$\tt{ = \dfrac{ 16 \: sin\bigg( \dfrac{2 \pi}{15} \bigg) \: cos \bigg( \dfrac{2 \pi}{15} \bigg) \:cos \bigg( \dfrac{4 \pi}{15} \bigg) \:cos \bigg( \dfrac{8 \pi}{15} \bigg) \:cos \bigg( \dfrac{14 \pi}{15} \bigg) }{ sin \bigg( \dfrac{2 \pi}{15} \bigg) }}$

$\tt{ = \dfrac{ 8 \cdot2 \: sin\bigg( \dfrac{2 \pi}{15} \bigg) \: cos \bigg( \dfrac{2 \pi}{15} \bigg) \cdot \: cos \bigg( \dfrac{4 \pi}{15} \bigg) \:cos \bigg( \dfrac{8 \pi}{15} \bigg) \:cos \bigg( \dfrac{14 \pi}{15} \bigg) }{ sin \bigg( \dfrac{2 \pi}{15} \bigg) }}$

$\tt{ = \dfrac{ 8 \: sin\bigg( \dfrac{4\pi}{15} \bigg) \: cos \bigg( \dfrac{4 \pi}{15} \bigg) \:cos \bigg( \dfrac{8 \pi}{15} \bigg) \:cos \bigg( \dfrac{14 \pi}{15} \bigg) }{ sin \bigg( \dfrac{2 \pi}{15} \bigg) }}$

$\tt{ = \dfrac{ 4 \cdot2 \: sin\bigg( \dfrac{4\pi}{15} \bigg) \: cos \bigg( \dfrac{4 \pi}{15} \bigg) \cdot \:cos \bigg( \dfrac{8 \pi}{15} \bigg) \:cos \bigg( \dfrac{14 \pi}{15} \bigg) }{ sin \bigg( \dfrac{2 \pi}{15} \bigg) }}$

$\tt{ = \dfrac{ 4 \: sin\bigg( \dfrac{8\pi}{15} \bigg) \:cos \bigg( \dfrac{8 \pi}{15} \bigg) \:cos \bigg( \dfrac{14 \pi}{15} \bigg) }{ sin \bigg( \dfrac{2 \pi}{15} \bigg) }}$

$\tt{ = \dfrac{ 2 \cdot 2\: sin\bigg( \dfrac{8\pi}{15} \bigg) \:cos \bigg( \dfrac{8 \pi}{15} \bigg) \cdot\:cos \bigg( \dfrac{14 \pi}{15} \bigg) }{ sin \bigg( \dfrac{2 \pi}{15} \bigg) }}$

$\tt{ = \dfrac{ 2 \: sin\bigg( \dfrac{16\pi}{15} \bigg) \:cos \bigg( \dfrac{14 \pi}{15} \bigg) }{ sin \bigg( \dfrac{2 \pi}{15} \bigg) }}$

$\tt{ = \dfrac{ 2 \: sin\bigg( \pi + \dfrac{\pi}{15} \bigg) \:cos \bigg( \pi - \dfrac{ \pi}{15} \bigg) }{ 2 \: sin \bigg( \dfrac{ \pi}{15} \bigg) \: cos \bigg( \dfrac{ \pi}{15} \bigg) }}$

$\tt{ = \dfrac{ 2 \: \cdot \bigg \{ - sin\bigg( \dfrac{\pi}{15} \bigg) \bigg \} \cdot \bigg \{ - cos \bigg( \dfrac{ \pi}{15} \bigg) \bigg \} }{ 2 \: sin \bigg( \dfrac{ \pi}{15} \bigg) \: cos \bigg( \dfrac{ \pi}{15} \bigg) }}$

$\tt{ = \dfrac{ 2 \: sin\bigg( \dfrac{\pi}{15} \bigg) \:cos \bigg( \dfrac{ \pi}{15} \bigg) }{ 2 \: sin \bigg( \dfrac{ \pi}{15} \bigg) \: cos \bigg( \dfrac{ \pi}{15} \bigg) }}$

$= \sf{1}$

Answer 2

${\boxed{\boxed{\begin{array}{cc}\bf \:L.H.S = \rm \: 16 \cos \frac{2\pi}{15} \cos\frac{4\pi}{15} \cos \frac{8\pi}{15} \cos \frac{14\pi}{15} \: \\ \\ \rm =16 \cos \frac{2\pi}{15} \cos2.\frac{2\pi}{15} \cos 4.\frac{2\pi}{15} \cos 7.\frac{2\pi}{15} \\ \\ \orange{ \bf \: let \: \: \: \theta = \frac{2\pi}{15} } \\ \\ \rm = 16 \: cos \theta \:. cos2 \theta.cos4 \theta.cos7 \theta \\ \\ \rm = \frac{1}{sin \theta} \times 8 \times (2 \: sin \theta.cos \theta).cos4 \theta.cos7 \theta \\ \\ \orange{ \boxed{ \begin{array}{cc} \sf \: we \: know \: that \\ \rm \: sin2x =2 sin \: x \: cos \: x \end{array}}} \\ \\ \rm = \frac{8}{sin \theta} \times sin2 \theta.cos4 \theta.cos7 \theta \\ \\ \rm = \frac{4}{sin \theta} \times (2 \: sin2 \theta.cos 2\theta ).cos4 \theta.cos7 \theta \\ \\ \rm = \frac{4}{sin \theta} \times sin4 \theta.cos 4\theta.cos7 \theta \\ \\ \rm = \frac{2}{sin \theta} \times (2 \: sin4 \theta.4cos \theta).cos7 \theta \\ \\ \rm = \frac{2}{sin \theta} \times sin8 \theta.cos7 \theta \\ \\ \end{array}}}}$

${\boxed{\boxed{\begin{array}{cc}{\orange{{\boxed{\begin{array}{cc}\bf \: we \: know \: that : \\ \\ \rm \: 2 \: sin \: x \: cos \: y= sin(x + y) + sin(x - y)\end{array}}}} }\\ \\ \rm = \frac{1}{sin \theta} \{sin(8 \theta + 7 \theta) + sin(8 \theta - 7 \theta) \} \\ \\ \rm = \frac{1}{sin \theta} \times (sin15 \theta + sin \theta) \\ \\ \rm = \frac{sin15 \theta}{sin \theta} + \frac{sin \theta}{sin \theta} \\ \\ \rm = \frac{1}{sin \theta} \times sin \: \cancel{15 }\times \frac{2\pi}{ \cancel{15}} + 1 \\ \\ \rm = \frac{1}{sin \theta} \times sin2\pi + 1 \\ \\ \rm = \frac{1}{sin \theta} \times 0 + 1 \\ \\ \orange{ \boxed{ \because \: \rm \: sin \: 2\pi = 0}} \\ \\ = 0 + 1 \\ \\ = 1 \\ \\ = \bf \: R.H.S\end{array}}}}$

Hence,

$\orange{\boxed{\boxed{\begin{array}{cc}\bf \: 16 \cos \frac{2\pi}{15} \cos\frac{4\pi}{15} \cos \frac{8\pi}{15} \cos \frac{14\pi}{15} = 1\end{array}}}}$

L.H.S = R.H.S _____(Proved)

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