Question

${3 \bigg[ { \sin^{4} \bigg( \frac{3\pi}{2} - \alpha \bigg) } + \sin^{4} (3\pi - \alpha ) \bigg] - 2 \bigg [ \sin^{6} \bigg( \frac{\pi}{2} + \alpha \bigg) + { \sin}^{6} (5\pi - \alpha ) \bigg]}$​

Answer 1

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

Given expression is

$\rm :\longmapsto\:{3 \bigg[ { \sin^{4} \bigg( \frac{3\pi}{2} - \alpha \bigg) } + \sin^{4} (3\pi - \alpha ) \bigg] - 2 \bigg [ \sin^{6} \bigg( \frac{\pi}{2} + \alpha \bigg) + { \sin}^{6} (5\pi - \alpha ) \bigg]}$

So, Let's evaluate first term of above expression.

Consider,

$\rm :\longmapsto\:{3 \bigg[ { \sin^{4} \bigg( \dfrac{3\pi}{2} - \alpha \bigg) } + \sin^{4} (3\pi - \alpha ) \bigg]}$

We know,

$\purple{\rm :\longmapsto\:\boxed{\tt{ sin\bigg(\dfrac{3\pi}{2} - x \bigg) = - cosx}}}$

So, using this we get

$\rm \:  =  \: 3\bigg[ {( - cos \alpha )}^{4} + {[sin(2\pi + \pi - \alpha )}^{4}\bigg]$

$\rm \:  =  \: 3\bigg[ {(cos \alpha )}^{4} + {[sin( \pi - \alpha )]}^{4}\bigg]$

$\rm \:  =  \: 3\bigg[ {(cos \alpha )}^{4} + {[sin(\alpha )]}^{4}\bigg]$

$\rm \:  =  \: 3\bigg[ {( {sin}^{2} \alpha )}^{2} + {( {cos}^{2} \alpha)}^{2}\bigg]$

We know,

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

So, using this identity, we get

$\rm \:  =  \: 3\bigg[ {( {sin}^{2} \alpha + {cos}^{2} \alpha )}^{2} - 2sin \alpha cos \alpha \bigg]$

$\rm \:  =  \: 3\bigg[ {(1 )}^{2} - 2sin \alpha cos \alpha \bigg]$

$\rm \:  =  \: 3\bigg[ 1 - 2sin \alpha cos \alpha \bigg]$

Now, Let's evaluate second term of the expression

Consider,

$\rm :\longmapsto\:{2 \bigg [ \sin^{6} \bigg( \frac{\pi}{2} + \alpha \bigg) + { \sin}^{6} (5\pi - \alpha ) \bigg]}$

We know,

$\purple{\rm :\longmapsto\:\boxed{\tt{ sin\bigg(\dfrac{\pi}{2} + x \bigg) = cosx}}}$

So, using this identity, we get

$\rm \:  =  \: 2\bigg[ {cos}^{6} \alpha + {sin}^{6}(4\pi + \pi - \alpha )\bigg]$

$\rm \:  =  \: 2\bigg[ {cos}^{6} \alpha + {sin}^{6}( \pi - \alpha )\bigg]$

$\rm \:  =  \: 2\bigg[ {cos}^{6} \alpha + {sin}^{6}(\alpha )\bigg]$

$\rm \:  =  \: 2\bigg[ {(cos \alpha )}^{6} + {(sin\alpha )}^{6}\bigg]$

$\rm \:  =  \: 2\bigg[ {( {sin}^{2} \alpha )}^{3} + {( {cos}^{2} \alpha)}^{3}\bigg]$

We know,

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

So, using this identity, we get

$\rm \:  =  \: 2\bigg[ {( {cos}^{2} \alpha + {sin}^{2} \alpha )}^{3} - 3 {sin}^{2} \alpha {cos}^{2} \alpha ( {cos}^{2} \alpha + {sin}^{2} \alpha )\bigg]$

$\rm \:  =  \: 2\bigg[ {(1)}^{3} - 3 {sin}^{2} \alpha {cos}^{2} \alpha ( 1)\bigg]$

$\rm \:  =  \: 2\bigg[1 - 3 {sin}^{2} \alpha {cos}^{2} \alpha \bigg]$

Now, Consider

$\rm :\longmapsto\:{3 \bigg[ { \sin^{4} \bigg( \frac{3\pi}{2} - \alpha \bigg) } + \sin^{4} (3\pi - \alpha ) \bigg] - 2 \bigg [ \sin^{6} \bigg( \frac{\pi}{2} + \alpha \bigg) + { \sin}^{6} (5\pi - \alpha ) \bigg]}$

$\rm \:  =  \: 3(1 - 2 {sin}^{2} \alpha {cos}^{2} \alpha ) - 2(1 - 3 {sin}^{2} \alpha {cos}^{2} \alpha )$

$\rm \:  =  \: 3 - 6{sin}^{2} \alpha {cos}^{2} \alpha - 2 + 6 {sin}^{2} \alpha {cos}^{2} \alpha$

$\rm \:  =  \: 1$

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MORE TO KNOW

Sign of Trigonometric ratios in Quadrants

sin (90°-θ)  =  cos θ

cos (90°-θ)  =  sin θ

tan (90°-θ)  =  cot θ

csc (90°-θ)  =  sec θ

sec (90°-θ)  =  csc θ

cot (90°-θ)  =  tan θ

sin (90°+θ)  =  cos θ

cos (90°+θ)  =  -sin θ

tan (90°+θ)  =  -cot θ

csc (90°+θ)  =  sec θ

sec (90°+θ)  =  -csc θ

cot (90°+θ)  =  -tan θ

sin (180°-θ)  =  sin θ

cos (180°-θ)  =  -cos θ

tan (180°-θ)  =  -tan θ

csc (180°-θ)  =  csc θ

sec (180°-θ)  =  -sec θ

cot (180°-θ)  =  -cot θ

sin (180°+θ)  =  -sin θ

cos (180°+θ)  =  -cos θ

tan (180°+θ)  =  tan θ

csc (180°+θ)  =  -csc θ

sec (180°+θ)  =  -sec θ

cot (180°+θ)  =  cot θ

sin (270°-θ)  =  -cos θ

cos (270°-θ)  =  -sin θ

tan (270°-θ)  =  cot θ

csc (270°-θ)  =  -sec θ

sec (270°-θ)  =  -csc θ

cot (270°-θ)  =  tan θ

sin (270°+θ)  =  -cos θ

cos (270°+θ)  =  sin θ

tan (270°+θ)  =  -cot θ

csc (270°+θ)  =  -sec θ

sec (270°+θ)  =  cos θ

cot (270°+θ)  =  -tan θ