已知cosa=cosx*siny,cosb=sinx*siny求证：sin^2a+sin^2b+sin^2y=2

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已知cosa=cosx*siny,cosb=sinx*siny求证：sin^2a+sin^2b+sin^2y=2

已知cosa=cosx*siny,cosb=sinx*siny求证：sin^2a+sin^2b+sin^2y=2
已知cosa=cosx*siny,cosb=sinx*siny
求证：sin^2a+sin^2b+sin^2y=2

已知cosa=cosx*siny,cosb=sinx*siny求证：sin^2a+sin^2b+sin^2y=2
因为 cosa=cosxsiny,cosb=sinxsiny
所以 (cosa)^2=(cosx)^2*(siny)^2,(cosb)^2=(sinx)^2*(siny)^2.
利用 (sina)^2=1-(cosa)^2,(sinb)^2=1-(cosb)^2 得到
(sina)^2+(sinb)^2+(siny)^2
=1-(cosx)^2*(siny)^2+1-(sinx)^2*(siny)^2+(siny)^2
=2-(siny)^2*[(cosx)^2+(sinx)^2]+(siny)^2
=2-(siny)^2+(siny)^2
=2
所证等式成立.

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