請問大大這兩題如何算 第一題我答案算-18/5可是不對 詳解.... 然後第二題也是 拜託了!?
回答 (3)
2017-11-06 8:12 pm
✔ 最佳答案
1.sin θ + cos θ = 2/3 , 兩邊平方得 :
1 + 2sinθcosθ = 4/9
sinθcosθ = - 5/18
sec θ + csc θ
= 1/cosθ + 1/sinθ
= ( sin θ + cos θ )/( sinθcosθ )
= (2/3) / (- 5/18)
= - 12/5 ..... Ans
2.
sin^4 θ + cos^4 θ
= ( sin² θ + cos² θ )² - 2( sinθcosθ )²
= 1 - 2(- 5/18)²
= 274/324 ..... Ans
2017-11-06 8:19 pm
1
Sol
Sinθ+Cosθ=2/3
Sin^2θ+2SinθCosθ+Cos^2θ=4/9
2SinθCosθ=－5/9
SinθCosθ=－5/18
Secθ+Cscθ
=Sinθ/(SinθCosθ)+Cosθ/(SinθCosθ)
=(2/3)/(－5/18)
=－(2/3)*(18/5)
=－12/5
2
Sin^4θ+Cos^4θ
=Sin^4θ+2Sin^2θCos^2θ+Cos^4θ－2Sin^2 θCos^2 θ
=(Sin^2θ+Cos^2θ)^2－2*(25/324)
=1－25/162
=137/162
Sol
Sinθ+Cosθ=2/3
Sin^2θ+2SinθCosθ+Cos^2θ=4/9
2SinθCosθ=－5/9
SinθCosθ=－5/18
Secθ+Cscθ
=Sinθ/(SinθCosθ)+Cosθ/(SinθCosθ)
=(2/3)/(－5/18)
=－(2/3)*(18/5)
=－12/5
2
Sin^4θ+Cos^4θ
=Sin^4θ+2Sin^2θCos^2θ+Cos^4θ－2Sin^2 θCos^2 θ
=(Sin^2θ+Cos^2θ)^2－2*(25/324)
=1－25/162
=137/162
2017-11-06 8:07 pm
(1) (sin θ + cos θ)² = sin²θ + cos²θ + 2 sin θ cos θ
=> (⅔)² = 1 + 2 sin θ cos θ
=> 2 sin θ cos θ = (⅔)² - 1 = 5/9
∴ sin θ cos θ = 5/18
1 1
sec θ + csc θ = ------- + ------
cos θ sin θ
sin θ + cos θ
= ------------------
cos θ sin θ
2/3
= ------
5/18
= (2/3)(18/5)
= 12/5
=> (⅔)² = 1 + 2 sin θ cos θ
=> 2 sin θ cos θ = (⅔)² - 1 = 5/9
∴ sin θ cos θ = 5/18
1 1
sec θ + csc θ = ------- + ------
cos θ sin θ
sin θ + cos θ
= ------------------
cos θ sin θ
2/3
= ------
5/18
= (2/3)(18/5)
= 12/5
收錄日期: 2021-04-30 22:36:16
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20171106102825AAuW2En