已知函数f(x)=2sin(x/3-派/6),x属于R一求f(5派/4)的值二设a,b属于[0,派/2],f(3a+派/2)=10/13,f(3b+2派)=6/5,求cos(a+b)的值

题目

已知函数f(x)=2sin(x/3-派/6),x属于R一求f(5派/4)的值二设a,b属于[0,派/2],f(3a+派/2)=10/13,f(3b+2派)=6/5,求cos(a+b)的值

答案

一.f(5Π/4)=2sin(5Π/4/3-Π/6)
=2sin(Π/4)= 根号2
二,f(3a+Π/2)=2sin(（3a+Π/2）/3-Π/6)=2sin(a)=10/13
sin(a)=5/13 cosa=12/13
f(3b+2Π)=2sin(（3b+2Π）/3-Π/6)=2sin(b+Π/2)=2cos(b)=6/5
cos(b)=3/5 sinb=4/5
cos(a+b)=cosa*cosb-sin(a)*sinb
=12/13*3/5+5/13*4/5
=56/65