Trigonometry

👨🏻‍🏫 學術台數學

kwan

2007-02-08 4:07 am

Express the form to rsin(θ-β)

3cosθ-sinθ

更新1:

r 0 , 0

更新2:

r 0 , 0

更新3:

要計算β的值

更新4:

但答案是 sq.root10 sin ((θ-251.6)

更新5:

但是別忘記β是在quadrant 3,我們應該加180度。

回答 (2)

Amby

2007-02-08 4:16 am

3cosθ-sinθ = rsin(θ-β)
= rsinθcosβ - rcosθsinβ
so rcosβ = -1, rsinβ = -3
r²cos²β + r²sin²β = (-1)² + (-3)²
r² = 10
r = -√10
rsinβ/rcosβ = 3
tanβ = 3
β = 71.57 °

so 3cosθ-sinθ = -√10sin(θ-71.57 °)

2007-02-07 20:17:02 補充：
β is corr to 4 s.f.

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myisland8132

2007-02-08 4:13 am

3cosθ-sinθ
=√10[(3/√10)(cosθ)-(1/√10)(sinθ)]
=√10[(sinβ)(cosθ)-(cosβ)(sinθ)] where sinβ=3/√10
=-√10[(cosβ)(sinθ)-(cosθ)(sinβ)]
=-√10sin(θ-β)

2007-02-07 20:26:45 補充：
sinβ=3/√10β=71.565計不到你的答案

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