Question

高一数学：已知a、B为锐角，cosa=1/7，sin（a+B）=5√3/14，求角B的值.

Answer 1

解：∵ α，β为锐角，
∴ sinα>0 cosβ>0 sinβ>0
∵ cosα=1/7
∴ sinα=√(1-cos²α)=√(1 - 1/7²)= (4√3)/7
∵ sin(α+β)= sinαcosβ+cosαsinβ
= (4√3)/7cosβ+(1/7)sinβ = (4√3)/7√(1-sin²β) +(1/7)sinβ
即：(4√3)/7√(1-sin²β) +(1/7)sinβ = 5√3/14
(4√3)/7√(1-sin²β) = 5√3/14 -(1/7)sinβ
两边平方
(48/49)(1-sin²β)= 75/196 - (5√3)/49sinβ + (1/49)sin²β
sin²β- (5√3)/49sinβ- 117/196 = 0
(sinβ- (5√3)/98)² = 5808/9604
sinβ- (5√3)/98 = (44√3)/98
sinβ = (49√3)/98 = (√3)/2
β = 60°

Answer 2

因为a为锐角，则有：
sina=√[1-(cosb)^2]=√[1-(1/7)^2]=4√3/7
cos(a+b)=±√{1-[sin(a+b)]^2}=±√{1-[5√3/14]^2}=±11/14
cosb=cos[(a+b)-a]
=cos(a+b)cosa +sin(a+b)sina
1.cos(a+b)取正值：
cosb=11/14×1/7+5√3/14×4√3/7=71/98=0.7245
所以，b=43.6°
2.cos(a+b)取负值：
cosb=-11/14×1/7+5√3/14×4√3/7=1/2
所以，b=60°

Answer 3

sina=√[1-(cosb)^2]=√[1-(1/7)^2]=4√3/7
cos(a+b)=±√{1-[sin(a+b)]^2}=±√{1-[5√3/14]^2}=±11/14
cosb=cos[(a+b)-a]
=cos(a+b)cosa +sin(a+b)sina
cos(a+b)取负值：
cosb=-11/14×1/7+5√3/14×4√3/7=1/2
b=60°