三角函数之求值问题
问题描述：
三角函数之求值问题
sin(pai/6-α)=1/3则cos(2pai/3+2α)= (pai为圆周率）
答
cos(2π/3+2α)
=-cos[π-(2π/3+2α)]
=-cos(π/3-2α)
=-cos[2(π/6-α)]
=-{1-2[sin(π/6-α)]^2}
=-(1-2/9)
=-7/9