cos(x)*cos(π/6-x)=1/2*[cos(π/6)+cos(2x-π/6)]是怎么导出来的?

问题描述:

cos(x)*cos(π/6-x)=1/2*[cos(π/6)+cos(2x-π/6)]是怎么导出来的?

利用积化和差公式 cosa*cosb=1/2[cos(a+b)+cos(a-b)] ∴cos(x)*cos(π/6-x) =1/2*[cos(x+π/6-x)]*[cos(x-(π/6-x))] =1/2*[cos(π/6)+cos(2x-π/6)] PS:积化和差公式sinαsinβ=-1/2[cos(α+β)-cos(α-β)] cosαcosβ=1/2[cos(α+β)+cos(α-β)] sinαcosβ=1/2[sin(α+β)+sin(α-β)] cosαsinβ=1/2[sin(α+β)-sin(α-β)] 和差化积公式:sinθ+sinφ=2sin(θ/2+θ/2)cos(θ/2-φ/2) sinθ-sinφ=2cos(θ/2+φ/2)sin(θ/2-φ/2) cosθ+cosφ=2cos(θ/2+φ/2)cos(θ/2-φ/2) cosθ-cosφ=-2sin(θ/2+φ/2)sin(θ/2-φ/2)