已知sin(x-π/6)=根号3/3求3sin2x+根号3cos2x

Question

已知sin(x-π/6)=根号3/3 求3sin2x+根号3cos2x

Answer 1

3sin2x+√ 3cos2x=2√ 3sin(2x+π/6)
sin(x-π/6)=√ 3/3
所以cos(x-π/6)=±√6/3
所以cos(2x-π/3)=
cos[2(x-π/6)]=cos^2(x-π/6)-sin^2(x-π/6)=1/3
而cos（2x-π/3)=-sin(2x+π/6)
所以3sin2x+√ 3cos2x=2√ 3sin(2x+π/6)=-2√ 3cos（2x-π/3)=-2√ 3/3