紧急求助：设a为锐角，若cos(a+π⼀6）=4⼀5，则sin(2a+π⼀12）的值为多少？（要详细过程）

解：
a是锐角
π/2cos(a+π/6)=4/5,
sin(a+π/6)=3/5
sin(2a+π/3)=2sin(a+π/6)cos(a+π/6)=2*(4/5)*(3/5)=24/25
cos(2a+π/3)=2cos²(a+π/6)-1=2*(4/5)²-1=7/25

sin(2a+π/12)
=sin[(2a+π/3)-π/4]
=sin(2a+π/3)cos(π/4)-cos(2a+π/3)cos(π/4)
=(24/25)*(√2/2)-(7/25)*(√2/2)
=17√2/50

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

五十分之十七倍根号二

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