Question

prove that Sin 5x =Sinx -20 Sin³x+16⁵x​

Answer 1

Appropriate Question :-

Prove that

$\rm \:sin5x = \:5sinx - {20sin}^{3}x+ {16sin}^{5}x$

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

Consider,

$\rm :\longmapsto\:sin5x$

can be rewritten as

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

We know,

$\boxed{ \bf{ \: sin(x + y) = sinxcosy +siny cosx}}$

So, using this, we get

$\rm \: = \:sin3xcos2x + sin2xcos3x$

We know,

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

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

$\boxed{ \bf{ \: sin2x = 2sinxcosx}}$

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

So, using these Identities, we get

$\rm \: =[3sinx - {4sin}^{3}x](1 - {2sin}^{2}x) + (2sinxcosx)[ {4cos}^{3}x - 3cosx]$

$\rm \: =[3sinx - {4sin}^{3}x](1 - {2sin}^{2}x) + 2sinxcos^{2} x[ {4cos}^{2}x - 3]$

$\rm \: =[3sinx - {4sin}^{3}x](1 - {2sin}^{2}x) + 2sinx[1 - sin^{2} x][ 4(1 - {sin}^{2}x) - 3]$

$\rm \: =[3sinx - {4sin}^{3}x](1 - {2sin}^{2}x) + 2sinx[1 - sin^{2} x][ 4 - 4{sin}^{2}x - 3]$

$\rm \: =[3sinx - {4sin}^{3}x](1 - {2sin}^{2}x) + 2sinx[1 - sin^{2} x][ 1 - 4{sin}^{2}x]$

$\rm \: = \:3sinx - {6sin}^{3}x - {4sin}^{3}x + {8sin}^{5}x + (2sinx - 2 {sin}^{3}x)(1 - 4 {sin}^{2}x)$

$\rm \: = \:3sinx - {10sin}^{3}x+ {8sin}^{5}x +2sinx - {8sin}^{3}x - {2sin}^{3}x + {8sin}^{5}x$

$\rm \: = \:5sinx - {20sin}^{3}x+ {16sin}^{5}x$

Hence,

$\rm \: \boxed{ \bf{ \: sin5x = \:5sinx - {20sin}^{3}x+ {16sin}^{5}x }}$

Additional Information :-

$\boxed{ \bf{ \: sin(x - y) = sinxcosy - sinycosx}}$

$\boxed{ \bf{ \: cos(x + y) = cosxcosy - sinxsiny}}$

$\boxed{ \bf{ \: cos(x - y) = cosxcosy + sinxsiny}}$

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

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

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

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