Question

prove that 1-1/2sin2x = \frac{\sin ^{3} x+\cos ^{3} x}{\sin x+\cos x}​

Answer 1

Question :-

Prove that

$\rm \: 1 - \dfrac{1}{2}sin2x = \dfrac{ {sin}^{3}x + {cos}^{3}x}{sinx + cosx}$

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

Consider, RHS

$\rm \: \dfrac{ {sin}^{3}x + {cos}^{3}x}{sinx + cosx}$

We know,

$\boxed{\rm{  \: {x}^{3} + {y}^{3} = (x + y)( {x}^{2} + {y}^{2} - xy) \: }}$

So, using this identity, we get

$\rm \: =  \: \: \dfrac{ \cancel{(sinx + cosx)} \: ( {sin}^{2} x + {cos}^{2} x - sinx \: cosx)}{ \cancel{sinx + cosx}}$

$\rm \: =  \: {sin}^{2}x + {cos}^{2}x - sinx \: cosx$

We know,

$\boxed{\sf{  \:\rm \: {sin}^{2}x + {cos}^{2}x = 1 \: \: }}$

So, using this identity, we get

$\rm \: =  \: \: 1 - sinx \: cosx \:$

can be rewritten as

$\rm \: =  \: \: 1 - \dfrac{1}{2} (2 \: sinx \: cosx )\:$

$\rm \: =  \: \: 1 - \dfrac{1}{2} sin2x\:$

Hence,

$\rm\implies \:\boxed{\rm{  \:1 - \dfrac{1}{2}sin2x = \dfrac{ {sin}^{3}x + {cos}^{3}x}{sinx + cosx} \: }}$

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Additional Information :-

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

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

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

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

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

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

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

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

Answer 2

$\sf \purple{RHS:}$

$\frac{\sin ^{3} x+\cos ^{3} x}{\sin x+\cos x}$

$\blue{ \boxed{ {a}^{3} + {b}^{3} = (a + b)( {a}^{2} -ab + {b}^{2} )}}$

$\implies\frac{ \cancel{( \sin x + \cos x)}(\sin ^{2} x+\cos ^{2} x - \sin x \cdot \cos x)}{ \cancel{\sin x+\cos x}}$

$\implies\sin ^{2} x+\cos ^{2} x - \sin x \cdot \cos x$

$\blue{ \boxed{ \sin ^{2} x + \cos ^{2} = 1}}$

$\implies1- \frac{1}{2} ( 2\sin x \cdot \cos x)$

$\blue{ \boxed{2 \sin(x) \cos(x) = \sin(2x) }}$

$\implies1- \frac{1}{2} \sin 2x$

$\orange{ \sf since \: \red{LHS=RHS}}$

$\sf \color{grey}{hence \: proved}$

HOPE IT HELPS!!