Question

Sin5A=16sinpower5A-20sincubeA+5sinA

Answer 1

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

Consider,

$\rm :\longmapsto\:sin5A$

$\rm \: = \:sin(3A + 2A)$

We know,

$\boxed{ \tt{ \: sin(x + y) = sinxcosy + sinycosx \: \: }}$

So, using this

$\rm \: = \:sin3Acos2A + sin2Acos3A$

We know,

$\boxed{ \tt{ \: sin2A = 2 \: sinAcosA}}$

$\boxed{ \tt{ \: cos2A = 1 - {2sin}^{2}A \: }}$

$\boxed{ \tt{ \: sin3A = 3sinA - {4sin}^{3} A \: }}$

$\boxed{ \tt{ \: cos3A = {4cos}^{3}A - 3cosA \: }}$

So, on substituting these values, we get

$\rm = (3sinA - {4sin}^{3}A)(1 - 2 {sin}^{2}A) + 2sinAcosA( {4cos}^{3}A - 3cosA)$

$\rm = (3sinA - {4sin}^{3}A)(1 - 2 {sin}^{2}A) + 2sinA {cos}^{2}A( {4cos}^{2}A - 3)$

$\rm = (3sinA - {4sin}^{3}A)(1 - 2 {sin}^{2}A) + 2sinA {(1 - sin}^{2}A)[{4(1 - sin}^{2}A) - 3]$

$\rm = 3sinA - {6sin}^{3}A - {4sin}^{3}A + {8sin}^{5}A + (2sinA - {2sin}^{3}A)[4 - 4 {sin}^{2}A - 3]$

$\rm = 3sinA - {10sin}^{3}A + {8sin}^{5}A + (2sinA - {2sin}^{3}A)[1 - 4 {sin}^{2}A ]$

$\rm = 3sinA - {10sin}^{3}A + {8sin}^{5}A + 2sinA - {8sin}^{3}A - {2sin}^{3}A + {8sin}^{5}A$

$\rm = 5sinA - {20sin}^{3}A + {16sin}^{5}A$

Hence,

$\: \: \: \: \boxed{ \tt{ \: \rm sin5A = 5sinA - {20sin}^{3}A + {16sin}^{5}A \: \: }}$

More to know :-

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

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

$\boxed{ \tt{ \: sinx + siny = 2sin\bigg[\dfrac{x + y}{2} \bigg]cos\bigg[\dfrac{x - y}{2} \bigg]}}$

$\boxed{ \tt{ \: sinx - siny = 2cos\bigg[\dfrac{x + y}{2} \bigg]sin\bigg[\dfrac{x - y}{2} \bigg]}}$

$\boxed{ \tt{ \: cosx + cosy = 2cos\bigg[\dfrac{x + y}{2} \bigg]cos\bigg[\dfrac{x - y}{2} \bigg]}}$