如何证明Sinx+Cosx=根号2*Cos（x-π/2）

Sin(x)+Cos(x)
=√2[(1/√2)Sin(x)+(1/√2)Cos(x)]
=√2[Sin(π/4)Sin(x)+Cos(π/4)Cos(x)]
=√2Cos(x-π/4)

其中用到公式
Cos(x+y)=Cos(x)Cos(y)-Sin(x)Sin(y)
把y=-π/4,带入即得
Cos(x-π/4)=Cos(x)Cos(π/4)+Sin(x)Sin(π/4)
而Cos(π/4)=Sin(π/4)=1/√2

两边同时平方 得：1+2sinxcosx=2cos平方（x-π/2）
又：
cos（x-π/2）=sinx
1+2sinxcosx=2sin平方x
2sinxcosx=2sin平方x-1
sin2x=-cos2x
应该是正不出来

Cos（x-π/2）=sinx
(sinx+cosx)^2=2sinx

Sinx+Cosx=√2*[sinxsin(π/4)+cosxcos(π/4)]
=√2*[cos(x-π/4)]

sinx+cosx=根号2*sin(x+π/4)
根号2*cos(x-π/2)=根号2*sinx

也就是说，要证明sin(x+π/4)=sinx,好像不对啊
倒像是解方程

Cos（x-π/2）=sinx
(sinx+cosx)^2=2sinx