(2)一條複角題

Question

已知tanA = 1/2 及 tanB = 1/5，其中A和B均為銳角。不得計算A和B的值，試求sin(A+B)的值。

Answer 1

已知tanA = 1/2 及 tanB = 1/5，其中A和B均為銳角。不得計算A和B的值，試求sin(A+B)的值。
tanA = 1/2
sinA = 1/[(5)^1/2]
cosA = 2/[(5)^1/2]
tanB = 1/5
sinB = 1/[(26)^1/2]
cosB = 5/[(26)^1/2]
sin(A + B)
= sinAcosB + sinBcosA
= {1/[(5)^1/2]}．{2/[(5)^1/2]} + {1/[(26)^1/2]}．{5/[(26)^1/2]}
= (2/5) + (5/26)
= 77/130

Answer 2

tanA=1/2
tanB=1/5

tan(A+B)=(tanA+tanB)/(1-tanAtanB)
=(1/2+1/5)/(1-1/2 1/5)
=(7/10)/(9/10)
=7/9

sin(A+B)=tan(A+B)/sqrt(1+tan(A+B)^2)
=(7/9)/sqrt(1+(7/9)^2)
=7/sqrt(130)