cos(x)*cos(π/6-x)=1/2*[cos(π/6)+cos(2x-π/6)]是怎麼導出來的?

cos(x)*cos(π/6-x)=1/2*[cos(π/6)+cos(2x-π/6)]是怎麼導出來的?

利用積化和差公式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:積化和差公式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)]和差化積公式: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)