高一三角函数题的解法(2008湖北真题)
g(x)=cosx*f(sinx)+sinx*f(cosx)
= cosx *√[(1-sinx)/(1+sinx)]+sinx *√[(1-cosx)/(1+cosx)]
=cosx*√[(1-sinx)？/(1+sinx)(1-sinx)]+sinx *√[(1-cosx)？/(1+cosx) (1-cosx)]
=cosx*√[(1-sinx)？/(1-sin？x)]+sinx*√[(1-cosx)？/(1-cos？x)]
=cosx*√[(1-sinx)？/cos？x]+sinx*√[(1-cosx)？/罪？x]
= cosx * |(1-sinx)/cosx |+sinx * |(1-cosx)/sinx |
因为x∈(π，17π/12)，sinx < 0，cosx & lt0，所以
= cosx *(1-sinx)/(-cosx)+sinx *(1-cosx)/(-sinx)
= -(1-sinx)-(1-cosx)
= -(1-sinx)-(1-cosx)
= sinx+ cosx-2
= √2sin(x+π/4)-2