三角函数题目:已知:cosa=cosx*siny cosb=sinx*siny 求证:sin^2*a+sin^2*b+sin^2*ysin^2*a+sin^2*b+sin^2*y 应该是 sin^2a+sin^2b+sin^2y =2
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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*ysin^2*a+sin^2*b+sin^2*y 应该是 sin^2a+sin^2b+sin^2y =2
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
三角函数题目:已知:cosa=cosx*siny cosb=sinx*siny 求证:sin^2*a+sin^2*b+sin^2*ysin^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
这是证明题?
sin^2*a+sin^2*b+sin^2*y=2(sinacos+sinbcosb+sinycosy)只要求出sinacos+sinbcosb+sinycosy=1就行了
sin²a+sin²b+sin²y
=1-cos²a+1-cos²b+sin²y
=2-cos²xsin²y-sin²xsin²y+sin²y
=2
你要证它等于什么??