独步沧海
幼苗
共回答了17个问题采纳率:94.1% 举报
f(θ)=sin(3/π-2θ)+cos(π/3+2θ)
=sin(3/π)cos(2θ)-cos(π/3)sin(2θ)+cos(3/π)cos(2θ)-sin(π/3)sin(2θ)
= (1+√3)/2cos(2θ)- (1+√3)/2 sin(2θ)
= (1+√3)/2 [cos2θ-sin2θ]
=(1+√3)/2*√2cos(2θ-π/4)
所以 2θ-π/4=2kπ,即 θ=kπ-π/8(k∈Z),f(θ)的最大值为(√2+√6)/2
1年前
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