三角函數求值

1.由根與係數:

tanα + tanβ = 3

tanα*tanβ = 2


2.tan和角公式

tan(α+β) = (tanα + tanβ)/1-tanα*tanβ

= 3 / ( 1-2 ) = -1

= -3

3.三角恆等式

(tanθ)^2 = (secθ)^2 - 1

(-3)^2 = [sec(α+β)]^2 - 1

[sec(α+β)]^2 = 10

4.secθ*cosθ=1

[sec(α+β)]^2= [1/cos(α+β)]^2 = 10

[cos(α+β)]^2 = 1/10 #

Money
用公式導出易懂
(tanθ)^2 = (secθ)^2 - 1

知識長很強但直接用一般人無法領悟

Tanα，Tanβ為x^2－3x+2=0的兩根，求Cos^2(α+β)之值
Sol
Tanα+Tanβ=3，TanαTanβ=2
Tan(α+β)
=(Tanα+Tanβ)/(1－TanαTanβ)
=3/(1－2)
=－3
Cos^2(α+β)
=1/[1+Tan^2(α+β)]
=1/(1+9)
=1/10