已知函数f(x)=sin(x+7π/4)+cos(x-3π/4),x∈R (1)求f(x)的最小正周期和最小值
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f(x)=sin(x+7π/4)+cos(x-3π/4)
=sin(x+2π-π/4)+cos(x-π+π/4)
=sin(x-π/4)+cos[-(π-x-π/4)]
=sin(x-π/4)+cos(π-x-π/4)
=sin(x-π/4)-cos(x+π/4)
=sin(x-π/4)-cos(x+π/2-π/4)
=sin(x-π/4)-cos(π/2+x-π/4)
=sin(x-π/4)+sin(x-π/4)
=2sin(x-π/4)
T=2π/1=2π
-2<=2sin(x-π/4)<=2
所以函数f(x)为：-2
=sin(x+2π-π/4)+cos(x-π+π/4)
=sin(x-π/4)+cos[-(π-x-π/4)]
=sin(x-π/4)+cos(π-x-π/4)
=sin(x-π/4)-cos(x+π/4)
=sin(x-π/4)-cos(x+π/2-π/4)
=sin(x-π/4)-cos(π/2+x-π/4)
=sin(x-π/4)+sin(x-π/4)
=2sin(x-π/4)
T=2π/1=2π
-2<=2sin(x-π/4)<=2
所以函数f(x)为：-2
参考资料： http://zhidao.baidu.com/question/285944638.html
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化简最终结果为：f(x)=根号2sinx,所以T=2派，最小值为-根号2
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