三角比,教下我(3)

a. √3sinθ = cosθ

√3sinθ/√3 = cosθ/√3

sinθ = cosθ/√3

sinθ/cosθ = cosθ/√3cosθ

tanθ = 1/√3

θ = 30°


b. 2cos3θ = 1

cos3θ = 1/2

3θ = 60°

θ = 20°



c. sinθ = 2/5

sin²θ = (2/5)² = 4/25

1 - sin²θ = 1 - 4/25 = 21/25

By definition, cos²θ = 1 - sin²θ

So, cos²θ = 21/25

cosθ = ±√(21/25) = √21/5 or -√21/5


When cosθ = √21/5

tanθ = sinθ/cosθ = (2/5)/(√21/5) = 2/√21 = 2√21/21 (將答案有理化)

When cosθ = -√21/5

tanθ = sinθ/cosθ = (2/5)/(-√21/5) = -2/√21 = -2√21/21 (將答案有理化)


a. 1 - (cosθ - sinθ)² = 2sinθcosθ

L.H.S.

= 1 - (cosθ - sinθ)²

= 1 - (cos²θ - 2sinθcosθ + sin²θ)

= 1 - [(cos²θ + sin²θ) - 2sinθcosθ]



By definition, cos²θ + sin²θ = 1

So, 1 - [(cos²θ + sin²θ) - 2sinθcosθ]

= 1 - (1 - 2sinθcosθ)

= 1 - 1 + 2sinθcosθ (負負得正)

= 2sinθcosθ

= R.H.S.


So, 1 - (cosθ - sinθ)² ≡ 2sinθcosθ

你個θ的範圍係幾多???
還是要找通解
我找通解好了 我計埋 0< θ <360

a)√3sinθ=cosθ (sinθ/cosθ=tanθ)
tanθ= 1/√3
θ=180n+30 度 n是整數 θ =30 ,210
b)2cos3θ=1
cos3θ=1/2
3θ=360n+60度 或3θ=360n-60度
θ=120n+20度 或 θ=120n-20度 θ =20,100,140,220,260,340
c) sinθ=2/5

cosθ=√(5²-2²) /5 或 cosθ=-√(5²-2²) /5
=√21 /5 = -√21 /5
tanθ=2/√(5²-2²) 或 tanθ= 2/ -√(5²-2²)
=2/√21 =2/ -√21
=(2√21) /21 = -(2√21) /21
d)左方=1-(cosθ-sinθ)²
=1-(cos²θ-2sinθcosθ+sin²θ)
=1-(1-2sinθcosθ)
=2sinθcosθ
=右方

三角比http://www.hkedcity.net/resources/common/video/res_video_download_index.phtml?res_cntr_id=16552