Given cos(42-θ)tan(42-θ)= cos(2θ=18),where θ is an acute angle.
(a)Find θby using the trigonometric identities.
(b)By using the result obtained in (a),find the value of 3 cos θsin(15+θ)tan20.
Please show steps.
麻煩幫幫忙啊，算了半天都算不出來……orz
[Maths]F.3 Trigonometry
2008-04-28 1:00 am
回答 (1)
2008-04-28 1:08 am
✔ 最佳答案
利用公式tanθ=sinθ/cosθ及sinθ＝cos(90-θ)(a)Find θby using the trigonometric identities.
cos(42-θ)tan(42-θ)= cos(2θ+18)
cos(42-θ)sin(42-θ)/cos(42-θ)= cos(2θ+18)
sin(42-θ)= cos(2θ+18)
cos(90-42+θ) = cos(2θ+18)
cos(48+θ) = cos(2θ+18)
48+θ =2θ+18
30=θ
(b) 3 cos θsin(15+θ)tan2θ
=3cos 30 sin(15+30)tan60
=3( (開方３)/2) (1/(開方２))(開方３)
=9/(2(開方２))
=9(開方２)/4
收錄日期: 2021-05-01 19:29:22
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