Prove the following trigonometric identities.

Solution:

Given that,

$a \cos ^{3} \theta+3 a \cos \theta \sin ^{2} \theta=m$

 

$a \sin ^{3} \theta+3 a \cos ^{2} \theta \sin \theta=n$

We have to prove $(m+n)^{\frac{2}{3}}+(m-n)^{\frac{2}{3}}=2 a^{\frac{2}{3}}$

Adding both the equations, we get

$m+n=a \cos ^{3} \theta+3 a \cos \theta \sin ^{2} \theta+a \sin ^{3} \theta+3 a \cos ^{2} \theta \sin \theta$

$=a\left(\cos ^{3} \theta+3 \cos ^{2} \theta \sin \theta+3 \cos \theta \sin ^{2} \theta+\sin ^{3} \theta\right)$

 

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

Also.

$m-n=a \cos ^{3} \theta+3 a \cos \theta \sin ^{2} \theta-\left(a \sin ^{3} \theta+3 a \cos ^{2} \theta \sin \theta\right)$

$=a\left(\cos ^{3} \theta-3 \cos ^{2} \theta \sin \theta+3 \cos \theta \sin ^{2} \theta-\sin ^{3} \theta\right)$

 

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

Therefore, we have

$(m+n)^{2 / 3}+(m-n)^{2 / 3}$

$=a^{2 / 3}(\cos \theta+\sin \theta)^{2}+a^{2 / 3}(\cos \theta-\sin \theta)^{2}$

 

$=a^{2 / 3}\left\{(\cos \theta+\sin \theta)^{2}+(\cos \theta-\sin \theta)^{2}\right\}$

$=a^{2 / 3}\left\{\left(\cos ^{2} \theta+2 \cos \theta \sin \theta+\sin ^{2} \theta\right)+\left(\cos ^{2} \theta-2 \cos \theta \sin \theta+\sin ^{2} \theta\right)\right\}$

$=a^{2 / 3}\left\{\left(\cos ^{2} \theta+\sin ^{2} \theta+2 \cos \theta \sin \theta\right)+\left(\cos ^{2} \theta+\sin ^{2} \theta-2 \cos \theta \sin \theta\right)\right\}$

$=a^{2 / 3}\{(1+2 \cos \theta \sin \theta)+(1-2 \cos \theta \sin \theta)\}$

$=a^{2 / 3}(1+2 \cos \theta \sin \theta+1-2 \cos \theta \sin \theta)$

$=2 a^{2 / 3}$

Hence proved.