f(x)=cos(2x+pai/4)+sin(2x+pai/4)求单调区间
问题描述:
f(x)=cos(2x+pai/4)+sin(2x+pai/4)求单调区间
本人预习中,没看到像这种关于两个三角函数组合求单调区间的解析,
答
cos(2x+pai/4)+sin(2x+pai/4)=√2*[√2/2cos(2x+pai/4)+√2/2sin(2x+pai/4)]
=√2*[sin45*cos(2x+pai/4)+cos45sin(2x+pai/4)]=√2sin(45+2x+45)=√2sin(x+90)=-√2cos2x
[k*pai,(k+1/2)*pai]