已知sin(x+π/6)=3分之根下3,怎么求sin^2(π/3-x)?

问题描述:

已知sin(x+π/6)=3分之根下3,怎么求sin^2(π/3-x)?

sin(π/3-x)=cos(π/2-(π/3-x))=cos(x+π/6)
所以
sin^2(π/3-x)
=cos^2(x+π/6)
=1-sin^2(x+π/6)
=1-1/3
=2/3

sin(π/3 - x)=sin[π/2 - (x + π/6)]=cos(x+π/6)
sin²(π/3 - x)=cos²(x+π/6) =1-1/3=2/3

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