三角函數求值
回答 (3)
2013-11-24 2:08 am
✔ 最佳答案
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 #
參考: money
2013-11-24 7:56 pm
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
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
2013-11-24 1:59 am
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
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
收錄日期: 2021-05-02 12:06:21
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20131123000016KK02345