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Question Number 192987 by Mingma last updated on 01/Jun/23

Answered by ajfour last updated on 01/Jun/23

let left vertical=2  2cos θ=ccos x  sin θ=csin x  4cos θcos θ=ccos (π−x−θ)  ..........        ...........       ...........  4cos^2 θ+ccos xcos θ=csin xsin θ  cos^2 x+((cos^2 x)/2)=sin^2 x  (3/2)(1−sin^2 x)=sin^2 x  sin x=(√(3/5))  x=tan^(−1) (√(2/3))

letleftvertical=22cosθ=ccosxsinθ=csinx4cosθcosθ=ccos(πxθ)................................4cos2θ+ccosxcosθ=csinxsinθcos2x+cos2x2=sin2x32(1sin2x)=sin2xsinx=35x=tan123

Commented by Mingma last updated on 01/Jun/23

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