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。
Trigonometry
回答 (2)
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.
= 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.
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
=√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
收錄日期: 2021-04-25 16:53:46
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