Question

a为锐角 cos(a+π/6)=4/5 求sin(2a+π/12)

Answer 1

解：
a是锐角
π/2<a+π/6<2π/3
cos(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

Answer 2

sin(2a+π/12)=cos[2(a+π/6)-π/4]=√2/2(cos[2(a+π/6)]+sin[2(a+π/6)])
因为cos(a+π/6)=4/5,所以sin(a+π/6)=3/5
cos[2(a+π/6)]=2[cos(a+π/6)]²-1=7/25,sin[2(a+π/6)]=cos(a+π/6)*sin(a+π/6)=24/25。
所以结果为√2/2*（31/25)=31√2/50.