已知cos(π/3-α)=根号3/3,求cos(2π/3+α)+cos2(7π/6+α)

网友回答

cos(π/3-α)=√3/3
cos(2π/3+α)+cos2(7π/6+α)
=-cos(π-(2π/3+α)) + cos(7π/3 +2α)
=-cos(π/3-α) + cos(π/3 +2α)
=-√3/3 - cos(2π/3-2α)
=-√3/3 - cos2(π/3-α)
=-√3/3 - 2[cos(π/3-α)]^2 +1
=-√3/3 - 2(1/3) + 1
=-√3/3 + 1/3
======以下答案可供参考======
供参考答案1:
∵cos(π/3-α)=根号3/3
cos(2π/3+α)
=cos[π-(π/3-α)]
=-cos(π/3-α)
=-√3/3
cos(7π/6+α)
=cos[π/2+(2π/3+α)]
=-sin(2π/3+α)
=-√(1-1/3)
=-√6/3
∴cos2(7π/6+α)
=2[cos(7π/6+α)]^2--1
=1/3∴cos(2π/3+α)+cos2(7π/6+α)
=-√3/3
+1/3