Wednesday, December 19, 2018

algebra precalculus - If costheta=fraccosalpha+cosbeta1+cosalphacosbeta, then prove that one value of tan(theta/2) is tan(alpha/2)tan(beta/2)




If cosθ=cosα+cosβ1+cosαcosβ

then prove that one of the values of tanθ2 is tanα2tanβ2.





I don't even know how to start this question. Pls help


Answer



I think i have found out how to approach it. I hope this proof is satifactory.



Using the half angle formula,
tanθ2=±1cosθ1+cosθeq.1


Evaluating 1cosθ1+cosθ first,
1cosθ1+cosθ=1(cosα+cosβ1+cosαcosβ)1+(cosα+cosβ1+cosαcosβ) [since cosθ=cosα+cosβ1+cosαcosβ]

=1+cosαcosβcosαcosβ1+cosαcosβ+cosα+cosβ

=(1cosα)(1cosβ)(1+cosα)(1+cosβ)



Substituting this value of 1cosθ1+cosθ into equation 1,
tanθ2=(1cosα)(1cosβ)(1+cosα)(1+cosβ)



=(1cosα)2(1cosβ)2(1cos2α)(1cos2β)

=±(1cosα)(1cosβ)sinαsinβ

Taking the positive value of tanθ2,
(1cosα)(1cosβ)sinαsinβ=4sin2α2sin2β24sinα2cosα2sinβ2cosβ2(using half angle and double angle formula)

=tanα2tanβ2



Therefore, tanα2tanβ2 is one of the values of tanθ2.


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