Question

Day 8,
$\begin{gathered}prove \: that \: \\ 16 \cos \frac{2\pi}{15} \cos\frac{4\pi}{15} \cos \frac{8\pi}{15} \cos \frac{14\pi}{15} = 1\end{gathered}$

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Answer 1

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

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

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)

Answer 2

$\large\underline{\sf{Solution-}}$

Consider LHS

$\rm :\longmapsto\:16cos\bigg[\dfrac{2\pi}{15} \bigg]cos\bigg[\dfrac{4\pi}{15} \bigg]cos\bigg[\dfrac{8\pi}{15} \bigg]cos\bigg[\dfrac{14\pi}{15} \bigg]$

can be rewritten as

$\rm = \: 16cos\bigg[\dfrac{2\pi}{15} \bigg]cos\bigg[\dfrac{4\pi}{15} \bigg]cos\bigg[\dfrac{8\pi}{15} \bigg]cos\bigg[\pi - \dfrac{\pi}{15} \bigg]$

We know,

$\red{\rm :\longmapsto\:\boxed{ \tt{ \: cos(\pi - x) = - cosx \: }}}$

So, using this, we get

$\rm = \: - 16cos\bigg[\dfrac{2\pi}{15} \bigg]cos\bigg[\dfrac{4\pi}{15} \bigg]cos\bigg[\dfrac{8\pi}{15} \bigg]cos\bigg[ \dfrac{\pi}{15} \bigg]$

$\rm = \: - 16cos\bigg[\dfrac{\pi}{15} \bigg]cos\bigg[\dfrac{2\pi}{15} \bigg]cos\bigg[\dfrac{4\pi}{15} \bigg]cos\bigg[ \dfrac{8\pi}{15} \bigg]$

can be further rewritten as

$\rm = \: - 16cos\bigg[\dfrac{\pi}{15} \bigg]cos\bigg[\dfrac{2\pi}{15} \bigg]cos\bigg[\dfrac{{2}^{2} \pi}{15} \bigg]cos\bigg[ \dfrac{ {2}^{3} \pi}{15} \bigg]$

Let we assume that

$\red{\rm :\longmapsto\:\boxed{ \tt{ \: \dfrac{\pi}{15} = x \: }}}$

So, above can be rewritten as

$\rm = \: - 16cosx \: cos2x \: cos {2}^{2}x \: cos {2}^{3}x \:$

We know that

$\boxed{ \tt{ \: cosxcos2xcos {2}^{2}xcos {2}^{3}x - - - cos {2}^{n}x = \dfrac{sin {2}^{n + 1}x }{ {2}^{n + 1}x} }}$

So, using this identity, we get

$\rm = \: - 16 \times \dfrac{sin {2}^{3 + 1}x}{ {2}^{3 + 1}sinx}$

$\rm = \: - 16 \times \dfrac{sin {2}^{4}x}{ {2}^{4}sinx}$

$\rm = \: - 16 \times \dfrac{sin 16x}{ 16sinx}$

$\rm = \: - \dfrac{sin 16x}{ sinx}$

Now, on substituting back the value of x, we get

$\rm = \: - \dfrac{sin \bigg[\dfrac{16\pi}{15} \bigg]}{ sin\bigg[\dfrac{\pi}{15} \bigg]}$

$\rm = \: - \dfrac{sin \bigg[\pi + \dfrac{\pi}{15} \bigg]}{ sin\bigg[\dfrac{\pi}{15} \bigg]}$

We know,

$\red{\rm :\longmapsto\:\boxed{ \tt{ \: sin(\pi - x) = sinx \: }}}$

So, using this identity, we get

$\rm = \: (- )( - )\dfrac{sin \bigg[ \dfrac{\pi}{15} \bigg]}{ sin\bigg[\dfrac{\pi}{15} \bigg]}$

$\rm = \: 1$

Hence,

$\red{\:\boxed{ \tt{ \: 16cos\bigg[\dfrac{2\pi}{15} \bigg]cos\bigg[\dfrac{4\pi}{15} \bigg]cos\bigg[\dfrac{8\pi}{15} \bigg]cos\bigg[\dfrac{14\pi}{15} \bigg] = 1 \: }}}$

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More to know :-

$\boxed{ \tt{ \: cosx \: cos\bigg[\dfrac{\pi}{3} - x\bigg] \: cos\bigg[\dfrac{\pi}{3} + x \bigg] = \frac{1}{4}cos3x \: }}$

$\boxed{ \tt{ \: sinx \: sin\bigg[\dfrac{\pi}{3} - x\bigg] \: sin\bigg[\dfrac{\pi}{3} + x \bigg] = \frac{1}{4}sin3x \: }}$

$\boxed{ \tt{ \: tanx \: tan\bigg[\dfrac{\pi}{3} - x\bigg] \: tan\bigg[\dfrac{\pi}{3} + x \bigg] = tan3x \: }}$

$\boxed{ \tt{ \: sin3x = 3sinx - {4sin}^{3}x \: }}$

$\boxed{ \tt{ \: cos3x = {4cos}^{3}x - 3cosx \: }}$

$\boxed{ \tt{ \: tan3x = \frac{3tanx - {tan}^{3} x}{1 - {3tan}^{2} x} \: }}$