InterviewSolution
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InterviewSolution
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Given `alpha+beta+gamma=pi,`prove that `sin^2alpha+sin^2beta-sin^2gamma=2sinalphasinbetacosgammadot` |
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Answer» `LHS = sin^(2)alpha+sin^(2)beta-sin^(2)gamma` `sin^(2)alpha+(sin^(2)beta-sin^(2)gamma)` `=sin^(2)alpha + sin(beta+gamma)sin(beta-gamma)` `=sin^(2)alpha + sin (pi -alpha)sin(beta - gamma)[because alpha+beta+gamma = pi]` `=sin^(2)alpha+sinalphasin(beta-gamma)` `=sin alpha [sinalpha + sin(beta-gamma)]` `=sin alpha[sin (pi-(beta+gamma))+sin(beta-gamma)]` `=sinalpha[sin(beta+gamma)+sin(beta-gamma)]` `=sin alpha[2 sin beta cos gamma]` `=2 sin alpha sin beta cos gamma = RHS` |
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