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)?

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Answer 1 · 2016-09-01T08:35:03

Given

$cos^2theta=(m^2-1)/3$

$=>3cos^2theta+1=m^2.....(1)$

Again given

$tan^3(theta/2)=tanalpha$

$=>tan^2(theta/2)=tan^(2/3)alpha...(2)$

We are to prove

$sin^(2/3)alpha+cos^(2/3)alpha=(2/m)^(2/3)$

Now expanding
$(sin^(2/3)alpha+cos^(2/3)alpha)^3$ we get

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

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

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

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

$=1+(3tan^(2/3)alpha(1+tan^(2/3)alpha))/(1+tan^2alpha)$

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

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

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

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

$color(red)("using relation (2)"->tan^2(theta/2)=tan^(2/3)alpha)$

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

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

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

$=1/((cos^2(theta/2)-sin^2(theta/2))^2+sin^2(theta/2)cos^2(theta/2))$

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

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

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

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

Hence proved

$sin^(2/3)alpha+cos^(2/3)alpha=(2/m)^(2/3)$