trigonometry 1

2011-08-26 8:10 am

回答 (1)

2011-08-26 12:19 pm
✔ 最佳答案
Q1.
For 180° < θ < 270°, sinθ < 0 and tanθ > 0.

sin²θ + cos²θ = 1
sin²θ + (-9/41)² = 1
sin²θ = 1600/1681
sinθ = 40/41 (rejected) or sinθ= -40/41

tanθ = sinθ/cosθ
tanθ = (-40/41)/(-9/41)
tanθ =40/9

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2.
In Quadrant IV, cosθ> 0 and tanθ < 0.

sin²θ + cos²θ = 1
(-0.6)² + cos²θ = 1
0.36 + cos²θ = 1
cos²θ = 0.64
cosθ = 0.8 or cosθ = -0.8 (rejected)

tanθ = sinθ/cosθ
tanθ = (-0.6)/(0.8)
tanθ =-0.75


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3.
In Quadrant II, sinθ > 0 and tanθ < 0.

sin²θ + cos²θ = 1
sin²θ + (-0.12)² = 1
sin²θ + 0.0144 = 1
sin²θ = 0.9856
sinθ = √0.9856 or sinθ = -√0.9856
sinθ = 0.993

tanθ = sinθ/cosθ
tanθ = (√0.9856)/(-0.12)
tanθ =-8.27


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5.
(a)
sin130°
= sin(180° - 50°)
= sin50°

(b)
sin310°
= sin(360° - 50°)
= -sin50°

(c)
sin410°
= sin(360° + 50°)
= sin50°


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6.
(a)
cos(-78°)
= cos78°

(b)
cos148°
= cos(180° - 32°)
= -cos32°

(c)
cos298°
= cos(360° - 62°)
= cos62°
參考: andrew


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