Question

已知a∈(0,π/4),b∈(π/4,3π/4),且cos(π/4-b)=4/5,sin(3π/4+a)=5/13，求cosb,cos(b+a)

Answer 1

∵a∈(0,π/4),b∈(π/4,3π/4),
∴-b∈(-3π/4,-π/4),(π/4-b)∈(-π/2,π/2），(3π/4+a)∈(3π/4,π）
cos(π/4-b)=4/5, sin(π/4-b)=3/5，sin(3π/4+a)=5/13,cos(3π/4+a)=-12/13
∴cosb=cos[π/4-(π/4-b)]=√2/2*4/5+√2/2*3/5=7√2/10,
cos(b+a)=sin[π/2+(b+a)]=sin[(3π/4+a)-(π/4-b)=5/13*4/5-(-12/13)*3/5=56/65
运算过程如上，具体数字自己再验算一下

Answer 2

∵ a∈(0,π/4),b∈(π/4,3π/4)
∴ π/4-b ∈ (-π/2, 0)
3π/4+a ∈ (3π/4, π)
∵ cos(π/4-b)= 4/5, sin (3π/4+a)= 5/13
∴ sin (π/4-b)= -3/5, cos(3π/4+a)= -12/13

这时，就有两种方法可以用了：
1.
利用三角函数的的和差公式，可以把4个算式写成：
cos(π/4-b) = cos(π/4) *cosb + sin(π/4)*sinb = 4/5, ①
sin (3π/4+a) = sin(3π/4)*cosa + cos(3π/4)*sina = 5/13 ②
sin (π/4-b) = sin(π/4)*cosb - cos(π/4)*sinb = -3/5, ③
cos(3π/4+a) = cos(3π/4)*cosa - sin(3π/4)*sina = -12/13 ④
①和③可以解得， sinb=7√2/10, cosb=√2/10
②和④可以解得， sina=7√2/26, cosa=17√2/26
∴ cosb=√2/10
cos(a+b)=cosa*cosb - sina*sinb
=√2/10×17√2/26 - 7√2/10×7√2/26
　 　 　 = -16/65

2.
cosb =cos[π/4-(π/4-b)]
=√2/2×4/5 + √2/2×(-3/5) =√2/10,

cos(b+a)=sin[π/2+(b+a)]
　 　 　 =sin[(3π/4+a)-(π/4-b)]
=5/13×4/5-(-12/13)×(-3/5)=-16/65
搞定，希望对你有所帮助~~~