Question

Show that sin pi/14.sin 3pi/14.sin 5pi/14 = 1/8​

Answer 1

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

Consider LHS

$\rm \: sin\dfrac{\pi}{14} \: sin\dfrac{3\pi}{14} \: sin\dfrac{5\pi}{14}$

We know,

$\boxed{\sf{ \: \: sinx = cos\bigg(\dfrac{\pi}{2} - x\bigg) \: \: }}$

So, using this result, the above can be rewritten as

$\rm \: = \: cos\bigg(\dfrac{\pi}{2} - \dfrac{\pi}{14} \bigg) \: cos\bigg(\dfrac{\pi}{2} - \dfrac{3\pi}{14} \bigg) \: cos\bigg(\dfrac{\pi}{2} - \dfrac{5\pi}{14} \bigg)$

$\rm \: = \: cos\bigg(\dfrac{7\pi - \pi}{14}\bigg)\: cos\bigg(\dfrac{7\pi - 3\pi}{14}\bigg)\: cos\bigg(\dfrac{7\pi - 5\pi}{14}\bigg)$

$\rm \: = \: cos\bigg(\dfrac{6\pi}{14}\bigg)\: cos\bigg(\dfrac{4\pi}{14}\bigg)\: cos\bigg(\dfrac{2\pi}{14}\bigg)$

$\rm \: = \: cos\bigg(\dfrac{3\pi}{7}\bigg)\: cos\bigg(\dfrac{2\pi}{7}\bigg)\: cos\bigg(\dfrac{\pi}{7}\bigg)$

can be further rewritten as

$\rm \: = \: cos\bigg(\pi - \dfrac{4\pi}{7}\bigg)\: cos\bigg(\dfrac{2\pi}{7}\bigg)\: cos\bigg(\dfrac{\pi}{7}\bigg)$

We know,

$\boxed{\sf{ \: \: cos(\pi - x) = - cosx \: }}$

So, using this result, we get

$\rm \: = \: - \: cos\bigg(\dfrac{4\pi}{7}\bigg)\: cos\bigg(\dfrac{2\pi}{7}\bigg)\: cos\bigg(\dfrac{\pi}{7}\bigg)$

can be re-arranged as

$\rm \: = \: - \: cos\bigg(\dfrac{\pi}{7}\bigg)\: cos\bigg(\dfrac{2\pi}{7}\bigg)\: cos\bigg(\dfrac{4\pi}{7}\bigg)$

$\rm \: = \: - \: cos\bigg(\dfrac{\pi}{7}\bigg)\: cos\bigg(2 \: \dfrac{\pi}{7}\bigg)\: cos\bigg( {2}^{2} \: \dfrac{\pi}{7}\bigg)$

We know,

$\boxed{\sf{ cosx \: cos2x \: cos {2}^{2}x \: - - - cos {2}^{n}x = \frac{ sin{2}^{n + 1}x }{ {2}^{n + 1} sinx} \: }}$

So, using this result, we get

$\rm \: = \: - \: \dfrac{sin {2}^{2 + 1} \dfrac{\pi}{7}}{ {2}^{2 + 1} sin\dfrac{\pi}{7}}$

$\rm \: = \: - \: \dfrac{sin {2}^{3} \dfrac{\pi}{7}}{ {2}^{3} sin\dfrac{\pi}{7}}$

$\rm \: = \: - \: \dfrac{sin \dfrac{8\pi}{7}}{ 8 sin\dfrac{\pi}{7}}$

$\rm \: = \: - \: \dfrac{sin\bigg(\pi + \dfrac{\pi}{7}\bigg)}{ 8 sin\dfrac{\pi}{7}}$

$\rm \: = \: - \: \dfrac{ - \: sin\bigg(\dfrac{\pi}{7}\bigg)}{ 8 sin\dfrac{\pi}{7}}$

$\rm \: = \: \dfrac{1}{8}$

Hence,

$\rm\implies \: \: \boxed{\tt{ \: \rm \: sin\dfrac{\pi}{14} \: sin\dfrac{3\pi}{14} \: sin\dfrac{5\pi}{14} = \frac{1}{8} \: \: }}$

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

ADDITIONAL INFORMATION

$\boxed{\tt{ sin2x = 2sinxcosx = \frac{2tanx}{1 + {tan}^{2} x} \: }}$

$\boxed{\tt{ cos2x = {cos}^{2}x - {sin}^{2}x = 1 - {2sin}^{2}x = {2cos}^{2}x - 1 \: }}$

$\boxed{\tt{ tan2x = \frac{2tanx}{1 - {tan}^{2} x} \: }}$

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

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

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

Answer 2

${\pmb{ \frak{\underline\red{Given:-}}}}$

$\: \: \: \: \: \: \bullet\tt { \: \: sin \: \frac{\pi}{14} } \: . \: \sin \frac{3\pi}{14} . \: sin \: \frac{5\pi}{14}$

${\pmb {\frak {\underline\red{Solution:-}}}}$

$\tt:\longmapsto\boxed{ \tt{Answer \: in \: the \: above \: attachment}}$

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

@Shivam

#BeBrainly