sin^2(θ+α) + sin^2(θ+β) – sin^2(α-β) = 2cos(α-β)sin(θ+α)sin(θ+β)

 
 
Prove that

\sin^2 (\theta + \alpha) + \sin^2 (\theta + \beta) - \sin^2 (\alpha - \beta) = 2 \, \cos (\alpha - \beta) \, \sin (\theta + \alpha) \, sin (\theta + \beta)

 
 
 
 
 

 
 
 
 
 
 
 
 
 
 
 
 
 
 

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About benvitalis

math grad - Interest: Number theory
This entry was posted in Number Puzzles and tagged . Bookmark the permalink.

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