已知sin(x-π/6)=根号3/3 求3sin2x+根号3cos2x
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已知sin(x-π/6)=根号3/3 求3sin2x+根号3cos2x
已知sin(x-π/6)=根号3/3 求3sin2x+根号3cos2x
已知sin(x-π/6)=根号3/3 求3sin2x+根号3cos2x
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
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