题目

设a=(0,π/3),β∈(π/6,π/2),且α,β满足5倍根号3sina+5cosa=8,根号2sinβ+根号6cosβ=2
求cos（a+π/6）的值
求cos（a+β）的值

提问时间：2020-12-10

答案

5倍根号3sina+5cosa=8
（根号3／2）sina＋（1／2）cosa=4/5
sin(π/3)sina+cos(π/3)cosa=4/5
cos(a-π/3)=4/5
因为a∈(0,π/3),所以a-π/3∈(-π/3,0)
所以sin(a-π/3)=-根号(1-(cos(a-π/3))^2)=-根号(1-(4/5)^)=-3/5
根号2sinβ+根号6cosβ=2
2根号2((1/2)sinb+(根号3/2)cosb)=2
(1/2)sinb+(根号3/2)cosb=根号2/2
sin(π/6)sinb+cos(π/6)cosb=根号2/2
cos(b-π/6)=根号2/2
因为β∈(π/6,π/2)
所以b-π/6∈(0,π/3)
所以sin(b-π/6)=根号(1-(cos(b-π/6))^2)=根号(1-(根号2/2)^2)=根号2/2
(1)cos(a+π/6)=cos(a-π/3+π/2)=cos(a-π/3)cos(π/2)-sin(a-π/3)sin(π/2)
=(4/5)*0-(-3/5)*1=3/5
因为a∈(0,π/3),所以a+π/6∈(π/6,π/2)
所以sin(a+π/6)=根号（1-(cos(a+π/6))^2)=根号(1-(3/5)^2)=4/5
(2)cos(a+b)=cos(a+π/6+b-π/6)=cos(a+π/6)cos(b-π/6)-sin(a+π/6)sin(b-π/6)
=3/5*(根号2/2)-4/5*(根号2/2)=- 根号2/10