1. Tardigrade
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  3. Mathematics
  4. Prove that sin2 α + sin2β -sin2 γ = 2 sin α sin β sin γ ,where α + β +γ = π

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Q. Prove that $sin^2 \alpha + sin^2\beta -sin^2 \gamma = 2 sin \alpha sin \beta sin \gamma ,where \alpha + \beta +\gamma = \pi$

IIT JEEIIT JEE 1978

Solution:

$LHS = sin^2\alpha + sin^2\beta - sin 2\gamma = sin^2\alpha + (sin^2\beta - sin^2\gamma) $
$= sin^2 \alpha + sin (\beta + \gamma) sin (\beta - \gamma)$
$= sin^2 \alpha + sin (\pi - \alpha)sin (\beta - \gamma) [\because \alpha + \beta + \gamma = \pi]$
$= sin^2 \alpha + sin \alpha sin (\beta - \gamma)$
$= sin \alpha [sin \alpha + sin (\beta - \gamma)]$
$= sin \alpha [sin (\pi-(\beta + \gamma))+sin (\beta - \gamma)]$
$= sin \alpha [sin (\beta + \gamma) + sin (\beta - \gamma)]$
$= sin \alpha [2 sin \beta cos \gamma ] = 2 sin \alpha sin \beta cos \gamma = RHS$