Question

prove that
$4sin\alpha* sin(\alpha +\frac{\pi }{3} )*sin(\alpha -\frac{\pi }{3} )=sin 3\alpha$

Answer 1

Appropriate Question :-

Prove that

$\rm :\longmapsto\:4 \: sin \alpha \: sin\bigg[ \alpha + \dfrac{\pi}{3} \bigg] \: sin\bigg[ \alpha - \dfrac{\pi}{3} \bigg] = - sin3\alpha$

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

Consider,

$\rm :\longmapsto\:4 \: sin \alpha \: sin\bigg[ \alpha + \dfrac{\pi}{3} \bigg] \: sin\bigg[ \alpha - \dfrac{\pi}{3} \bigg]$

We know,

$\pink{\boxed{ \tt{ \: sin(x + y) \: sin(x - y) = {sin}^{2}x - {sin}^{2}y \: }}}$

So, using this identity, we get

$\rm \: = \: 4 \: sin \alpha \bigg[ {sin}^{2}\alpha - {sin}^{2}\dfrac{\pi}{3} \bigg]$

We know,

$\boxed{ \tt{ \: sin\dfrac{\pi}{3} = \frac{ \sqrt{3} }{2} \: }}$

So, using this, we get

$\rm \: = \: 4 \: sin\alpha \bigg[ {sin}^{2}\alpha - {\bigg[\dfrac{ \sqrt{3} }{2} \bigg]}^{2} \bigg]$

$\rm \: = \: 4 \: sin\alpha \bigg[ {sin}^{2}\alpha - \dfrac{3}{4} \bigg]$

$\rm \: = \: 4 \: sin\alpha \bigg[\dfrac{4 {sin}^{2}\alpha - 3}{4} \bigg]$

$\rm \: = \: sin\alpha \: (4 {sin}^{2}\alpha - 3)$

$\rm \: = \: {4sin}^{3}\alpha - 3sin\alpha$

$\rm \: = \: - \: ( \: 3sin\alpha - {4sin}^{3} \alpha \: )$

$\rm \: = \: - \: sin3\alpha$

Hence,

$\boxed{ \tt{ \: \:4 \: sin \alpha \: sin\bigg[ \alpha + \dfrac{\pi}{3} \bigg] \: sin\bigg[ \alpha - \dfrac{\pi}{3} \bigg] = - sin3\alpha }}$

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Additional Information :-

$\boxed{ \tt{ \: sin2x = 2sinx \: cosx \: }}$

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

$\boxed{ \tt{ \: cos2x = \frac{1 - {tan}^{2} x}{1 + {tan}^{2}x } \: }}$

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

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

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

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

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

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