Question

高一数学，第九题，求解，

高一数学，第九题，求解，急在线等，最好有解析，

Answer 1

y=[cos(x-π/12)]^2-[sin(x+π/12)]^2
=1/2[1+cos2(x-π/12)]-1/2[1-cos2(x+π/12)]
=1/2cos(2x-π/6)+1/2cos(2x+π/6)
=1/2(cosπ/6cos2x+sin2xsinπ/6)+1/2(cosπ/6cos2x-sin2xsinπ/6)
=1/2(√3/2cos2x+1/2sin2x+1/2(√3/2cos2x-1/2sin2x)
=√3/4cos2x+1/4sin2x+√3/4cos2x-1/4sin2x
=√3/2cos2x
T=2π/2=π
f(x)=√3/2cos2x
因f(-x)=√3/2cos(-2x)=√3/2cos2x=f(x)
f(-x)=f(x)
所以y=[cos(x-π/12)]^2-[sin(x+π/12)]^2=√3/2cos2x是偶函数
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