球神不败佛
幼苗
共回答了17个问题采纳率:94.1% 举报
1.f(x)=2sinxcosx+cos2x=sin2x+cos2x=√2sin(2x+π/4)
f(a/2)=√2sin(a+π/4)=-√2/2
所以sin(a+π/4)=-1/2
因为a∈(0,π),所以a+π/4∈(0,5π/4),
所以a+π/4=7π/6=11π/12
2.sin(a+π/6)=cos[π/2-(a+π/6)]=cos(π/3-a)=1/3
cos(2π/3-2a)=cos2(π/3-a)=2cos^2(π/3-a)-1=2*(1/3)^2-1=-7/9
1年前
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