sin(6/π-α)=1/3,则cos(2π/3+2α)的值为

问题描述:

sin(6/π-α)=1/3,则cos(2π/3+2α)的值为

cos(2π/3+2α)= - cos(π-(2π/3+2α))=- cos(π/3-2α)=- cos(2α-π/3) (公式cos2α=1-2(sinα)^2)=- (1-2(sin(α-π/6))^2)∵sin(π/6-α)=1/3 ∴sin(α-π/6))^2=1/9故cos(2π/3+2α)= -(1-2*1/9) =-7/9...