1个回答
2014-02-23
展开全部
1-2sin^2(x+兀/8)=cos(2x+兀/4)
2sin(x+兀/8)cos(x+兀/8)=sin(2x+兀/4)
f(x)=cos(2x+兀/4)+sin(2x+兀/4)
=√2*[sin兀/4*cos(2x+兀/4)+cos兀/4*sin(2x+兀瞎渣/4)]
=√2*sin(2x+兀/2)
T=2兀/2=兀
f(x)的单调增区睁神缓间:
-兀/2/+2k兀≤悉模2x+兀/2≤兀/2+2k兀
-兀/2/8+K兀≤x≤0+k兀
k属于Z
2sin(x+兀/8)cos(x+兀/8)=sin(2x+兀/4)
f(x)=cos(2x+兀/4)+sin(2x+兀/4)
=√2*[sin兀/4*cos(2x+兀/4)+cos兀/4*sin(2x+兀瞎渣/4)]
=√2*sin(2x+兀/2)
T=2兀/2=兀
f(x)的单调增区睁神缓间:
-兀/2/+2k兀≤悉模2x+兀/2≤兀/2+2k兀
-兀/2/8+K兀≤x≤0+k兀
k属于Z
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