# If cos^2 θ = (m^2 - 1)/3 and tan^3 (θ/2) = tan α, then prove that sin^(2/3) α + cos^(2/3) α = (2/m)^(2/3)?

Sep 1, 2016

Given

${\cos}^{2} \theta = \frac{{m}^{2} - 1}{3}$

$\implies 3 {\cos}^{2} \theta + 1 = {m}^{2.} \ldots . \left(1\right)$

Again given

${\tan}^{3} \left(\frac{\theta}{2}\right) = \tan \alpha$

$\implies {\tan}^{2} \left(\frac{\theta}{2}\right) = {\tan}^{\frac{2}{3}} \alpha \ldots \left(2\right)$

We are to prove

${\sin}^{\frac{2}{3}} \alpha + {\cos}^{\frac{2}{3}} \alpha = {\left(\frac{2}{m}\right)}^{\frac{2}{3}}$

Now expanding
${\left({\sin}^{\frac{2}{3}} \alpha + {\cos}^{\frac{2}{3}} \alpha\right)}^{3}$ we get

${\left({\sin}^{\frac{2}{3}} \alpha + {\cos}^{\frac{2}{3}} \alpha\right)}^{3}$
$= \left({\sin}^{2} \alpha + {\cos}^{2} \alpha\right) + 3 {\sin}^{\frac{2}{3}} \alpha {\cos}^{\frac{2}{3}} \alpha \left({\sin}^{\frac{2}{3}} \alpha + {\cos}^{\frac{2}{3}} \alpha\right)$

$= 1 + 3 {\sin}^{\frac{2}{3}} \alpha {\cos}^{\frac{2}{3}} \alpha \left({\sin}^{\frac{2}{3}} \alpha + {\cos}^{\frac{2}{3}} \alpha\right)$

=1+(3sin^(2/3)alphacos^(2/3)alpha(sin^(2/3)alpha+cos^(2/3)alpha))/((sin^2alpha+cos^2alpha)

$= 1 + \frac{\frac{3 {\sin}^{\frac{2}{3}} \alpha {\cos}^{\frac{2}{3}} \alpha \left({\sin}^{\frac{2}{3}} \alpha + {\cos}^{\frac{2}{3}} \alpha\right)}{\cos} ^ 2 \alpha}{\frac{{\sin}^{2} \alpha + {\cos}^{2} \alpha}{\cos} ^ 2 \alpha}$

$= 1 + \frac{3 {\tan}^{\frac{2}{3}} \alpha \left(1 + {\tan}^{\frac{2}{3}} \alpha\right)}{1 + {\tan}^{2} \alpha}$

=1+(3tan^(2/3)alpha(1+tan^(2/3)alpha))/((1+tan^(2/3)alpha)(1-tan^(2/3)alpha+tan^(4/3)alpha)

$= 1 + \frac{3 {\tan}^{\frac{2}{3}} \alpha}{1 - {\tan}^{\frac{2}{3}} \alpha + {\tan}^{\frac{4}{3}} \alpha}$

$= \frac{1 - {\tan}^{\frac{2}{3}} \alpha + {\tan}^{\frac{4}{3}} \alpha + 3 {\tan}^{\frac{2}{3}} \alpha}{1 - {\tan}^{\frac{2}{3}} \alpha + {\tan}^{\frac{4}{3}} \alpha}$

$= {\left(1 + {\tan}^{\frac{2}{3}} \alpha\right)}^{2} / \left(1 - {\tan}^{\frac{2}{3}} \alpha + {\tan}^{\frac{4}{3}} \alpha\right)$

$\textcolor{red}{\text{using relation (2)} \to {\tan}^{2} \left(\frac{\theta}{2}\right) = {\tan}^{\frac{2}{3}} \alpha}$

$= {\left(1 + {\tan}^{2} \left(\frac{\theta}{2}\right)\right)}^{2} / \left(1 - {\tan}^{2} \left(\frac{\theta}{2}\right) + {\tan}^{4} \left(\frac{\theta}{2}\right)\right)$

=1/(cos^4(theta/2)(1-tan^2(theta/2)+tan^4(theta/2))

$= \frac{1}{{\cos}^{4} \left(\frac{\theta}{2}\right) - {\sin}^{2} \left(\frac{\theta}{2}\right) {\cos}^{2} \left(\frac{\theta}{2}\right) + {\sin}^{4} \left(\frac{\theta}{2}\right)}$

$= \frac{1}{{\left({\cos}^{2} \left(\frac{\theta}{2}\right) - {\sin}^{2} \left(\frac{\theta}{2}\right)\right)}^{2} + {\sin}^{2} \left(\frac{\theta}{2}\right) {\cos}^{2} \left(\frac{\theta}{2}\right)}$

=1/((cos^2theta+1/4*4*sin^2(theta/2)cos^2(theta/2))

=4/((4cos^2theta+sin^2theta)

=4/((4cos^2theta+1-cos^2theta)

=4/((3cos^2theta+1))=4/m^2=(2/m)^2->color(red)("by relation(1))"

Hence proved

${\sin}^{\frac{2}{3}} \alpha + {\cos}^{\frac{2}{3}} \alpha = {\left(\frac{2}{m}\right)}^{\frac{2}{3}}$