Question

if $\tt acos^3 θ+3acos\: θ\: sin^2 θ=m$ and $\tt asin^3 θ+3acos^2 θ \: sin\: θ=n$
then prove that,
$\tt (m+n)^{\frac{2}{3}} + (m-n)^{\frac{2}{3}}=2a^{\frac{2}{3}}$
​

Answer 1

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

Given that

$\rm :\longmapsto\:a {cos}^{3}\theta + 3acos\theta \: {sin}^{2}\theta = m$

and

$\rm :\longmapsto\:a {sin}^{3}\theta + 3asin\theta \: {cos}^{2}\theta = n$

Now, Consider

$\rm :\longmapsto\:m + n$

$\rm \:  =  \: a {cos}^{3}\theta + 3acos\theta \: {sin}^{2}\theta + a {sin}^{3}\theta + 3asin\theta \: {cos}^{2}\theta$

$\rm \:  =  \: a \bigg[{cos}^{3}\theta + 3cos\theta \: {sin}^{2}\theta + {sin}^{3}\theta + 3sin\theta \: {cos}^{2}\theta\bigg]$

can be rewritten as

$\rm \:  =  \: a \bigg[{cos}^{3}\theta + 3cos\theta \: {sin}^{2}\theta + 3cos\theta \: {sin}^{2}\theta + {cos}^{3} \theta \bigg]$

We know,

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

So, using this identity, we get

$\rm \:  =  \: a {(cos\theta + sin\theta )}^{3}$

$\rm \implies\:m + n  =  \: a {(cos\theta + sin\theta )}^{3}$

So,

$\rm :\longmapsto\: {\bigg(m + n \bigg) }^{\dfrac{2}{3} }$

$\rm \:  =  \: {\bigg(a {(cos\theta + sin\theta )}^{3} \bigg) }^{\dfrac{2}{3} }$

$\rm \:  =  \: {\bigg(a\bigg) }^{\dfrac{2}{3} } {(cos\theta + sin\theta )}^{2}$

$\rm \implies\:\boxed{ \tt{ \: {\bigg(m + n \bigg) }^{\dfrac{2}{3} } = {\bigg(a\bigg) }^{\dfrac{2}{3} } {(cos\theta + sin\theta )}^{2} \: }}$

Now, Consider

$\rm :\longmapsto\:m - n$

$\rm \:  =  \: a {cos}^{3}\theta + 3acos\theta \: {sin}^{2}\theta - a {sin}^{3}\theta - 3asin\theta \: {cos}^{2}\theta$

$\rm \:  =  \: a \bigg[{cos}^{3}\theta + 3cos\theta \: {sin}^{2}\theta - {sin}^{3}\theta - 3sin\theta \: {cos}^{2}\theta\bigg]$

can be rewritten as

$\rm \:  =  \: a \bigg[{cos}^{3}\theta + 3cos\theta \: {sin}^{2}\theta - 3sin\theta \: {cos}^{2}\theta -{sin}^{3} \theta \bigg]$

$\rm \:  =  \: a {(cos\theta - sin\theta )}^{3}$

$\rm \implies\:m - n  =  \: a {(cos\theta - sin\theta )}^{3}$

So,

$\rm :\longmapsto\: {\bigg(m - n \bigg) }^{\dfrac{2}{3} }$

$\rm \:  =  \: {\bigg(a {(cos\theta - sin\theta )}^{3} \bigg) }^{\dfrac{2}{3} }$

$\rm \:  =  \: {\bigg(a\bigg) }^{\dfrac{2}{3} } {(cos\theta - sin\theta )}^{2}$

$\rm \implies\:\boxed{ \tt{ \: {\bigg(m - n \bigg) }^{\dfrac{2}{3} } = {\bigg(a\bigg) }^{\dfrac{2}{3} } {(cos\theta - sin\theta )}^{2} \: }}$

Now, Consider

$\red{\rm :\longmapsto\:{\bigg(m + n \bigg) }^{\dfrac{2}{3} } + {\bigg(m - n \bigg) }^{\dfrac{2}{3} }}$

On substituting the values, evaluated above, we get

$\rm \:  =  \: {\bigg(a\bigg) }^{\dfrac{2}{3} } {(cos\theta + sin\theta )}^{2} + {\bigg(a\bigg) }^{\dfrac{2}{3} } {(cos\theta - sin\theta )}^{2}$

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

We know

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

So, using this, we get

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

$\rm \:  =  \: {\bigg(a\bigg) }^{\dfrac{2}{3} } \bigg[2(1 ) \bigg]$

$\rm \:  =  \: 2{\bigg(a\bigg) }^{\dfrac{2}{3} }$

Hence,

$\red{\rm :\longmapsto\:\boxed{ \tt{ \: {\bigg(m + n \bigg) }^{\dfrac{2}{3} } + {\bigg(m - n \bigg) }^{\dfrac{2}{3} }} =2 {\bigg(a\bigg) }^{\dfrac{2}{3} } \: }}$

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Additional Information:-

Relationship between sides and T ratios

sin θ = Opposite Side/Hypotenuse

cos θ = Adjacent Side/Hypotenuse

tan θ = Opposite Side/Adjacent Side

sec θ = Hypotenuse/Adjacent Side

cosec θ = Hypotenuse/Opposite Side

cot θ = Adjacent Side/Opposite Side

Reciprocal Identities

cosec θ = 1/sin θ

sec θ = 1/cos θ

cot θ = 1/tan θ

sin θ = 1/cosec θ

cos θ = 1/sec θ

tan θ = 1/cot θ

Co-function Identities

sin (90°−x) = cos x

cos (90°−x) = sin x

tan (90°−x) = cot x

cot (90°−x) = tan x

sec (90°−x) = cosec x

cosec (90°−x) = sec x

Fundamental Trigonometric Identities

sin²θ + cos²θ = 1

sec²θ - tan²θ = 1

cosec²θ - cot²θ = 1