Question

21题，求解

Answer 1

1、f(π/6 )=sin(πw/6+π/3 )=1
所以πw/6+π/3=2kπ +π/2 解得w=12k+1
所以当k=0时，w取最小正值，此时w=1
f(x)=sin(x+π/3 )
2、利用两角差的正切公式，和两角差的余弦公式.不需要求出具体的值，
a∈( π/6,2π/3),f(a)=sin(a+π/3)=3/5
a+π/3∈( π/2,π),，所以cos(a+π/3)=-4/5
.b∈(-5π/6,- π/3 ).f(b)= sin(b+π/3)=-4/5.
b+π/3∈( -π/2,0),，所以cos(a+π/3)=3/5

tana=tan[(a+π/3)-π/3]=[-3/4 -√3 ]/[1-3√3 /4]=. =(48+25√3)/11

cos(a-b)=cos[(a+π/3)-(b+π/3)]=cos(a+π/3)cos(b+π/3) +sin(a+π/3)sin(b+π/3)=. =-24/25