神马作文网三角函数题目：已知：cosa=cosx*siny cosb=sinx*siny 求证：sin^2*a+sin^2*b+s
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三角函数题目：已知：cosa=cosx*siny cosb=sinx*siny 求证：sin^2*a+sin^2*b+sin^2*y
sin^2*a+sin^2*b+sin^2*y 应该是 sin^2a+sin^2b+sin^2y =2
sin^2*a+sin^2*b+sin^2*y 应该是 sin^2a+sin^2b+sin^2y =2
解析：∵cosa=cosx*siny,cosb=sinx*siny,
∴cosx=cosa/siny,sinx=cosb/siny
则(cosx)^2+(sinx)^2
=(cosa)^2/(siny)^2+(cosb)^2/(siny)^2=1
即（cosa)^2+(cosb)^2=(siny)^2
1-(sina)^2+1-(sinb)^2=(siny)^2
∴(sina)^2+（Sinb)^2+(siny)^2=2
∴cosx=cosa/siny,sinx=cosb/siny
则(cosx)^2+(sinx)^2
=(cosa)^2/(siny)^2+(cosb)^2/(siny)^2=1
即（cosa)^2+(cosb)^2=(siny)^2
1-(sina)^2+1-(sinb)^2=(siny)^2
∴(sina)^2+（Sinb)^2+(siny)^2=2
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