设a为锐角,若cos(a+π/6)=4/5,则sin(2a+π/12)的值为多少?

问题描述:

设a为锐角,若cos(a+π/6)=4/5,则sin(2a+π/12)的值为多少?

五十分之十七倍根号二

a为锐角则 π/6sin(2a+π/3)=2sin(a+π/6)cos(a+π/6)=24/25
cos(2a+π/3)=2cos^(a+π/6)-1=7/25
sin(2a+π/12)=sin(2a+π/3-π/4)=√2/2(sin(2a+π/3)-cos(2a+π/3))=17√2/50

设b=a+π/6,sinb=3/5,sin2b=2sinbcosb=24/25,cos2b=7/25
sin(2a+π/12)=sin(2a+π/3-π/4)=sin(2b-π/4)=sin2bcosπ/4-cos2bsinπ/4=(17√2)/50

设b=a+π/6,sinb=3/5,sin2b=2sinbcosb=24/25,cos2b=7/25
sin(2a+π/12)=sin(2a+π/3-π/4)=sin(2b-π/4)=sin2bcosπ/4-cos2bsinπ/4=(17√2)/50