sin(x+π/3)=2/3,求sin(x-2π/3)+sin^2(π/6-x)的值

问题描述:

sin(x+π/3)=2/3,求sin(x-2π/3)+sin^2(π/6-x)的值

sin(x-2π/3)+sin^2(π/6-x)
=-sin[π+(x-2π/3)]+cos^2[π/2-(π/6-x)]
=-sin(x+π/3)+cos^2(x+π/3)
=-sin(x+π/3)+[1-sin^2(x+π/3)]
=-2/3+(1-4/9)
=-1/9