標題:
trigonometry 1
圖片參考：http://imgcld.yimg.com/8/n/HA00437808/o/701108190057213873445901.jpg thanks~~~~! 更新: answers: 1. 120,quad.II 2. -40, quad.IV 3. 230, quad.III 4. sinΘ=3/5, cosΘ=4/5, tanΘ=3/4 5. sinΘ=-1/(sqrt2), cosΘ=1/sqrt2, tanΘ=-1 6. sinΘ=1/2, cosΘ=-(sqrt3)/2, tanΘ=-1/sqrt3 7. sinΘ=0, cosΘ=-1, tanΘ=0
最佳解答:
Quadrant I =0<θ<90 Quadrant II =90<θ<180 Quadrant III=180<θ<270 Quadrant IV =270<θ<360 sinθ=y/r, cosθ=x/r, tanθ=y/x [r=PO] So, 1. θ=90+30=120[Quadrant II] 2. θ=360-320=40[Quadrant I] 3. θ=270-40=230[Quadrant III] 4. sinθ=3/5, cosθ=4/5, tan=3/4 5. sinθ=-1/√2, cosθ=1/√2, tan=-1 6. sinθ=1/2, cosθ=-√3/2, tan=-1/√3 7. sinθ=0, cosθ=-1, tan=0 2011-08-19 21:02:16 補充： CORRECTION: 2. 320-θ=360, θ=-40[Quadrant IV] sorry for the mistake...
其他解答:
trigonometry 1
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發問:圖片參考：http://imgcld.yimg.com/8/n/HA00437808/o/701108190057213873445901.jpg thanks~~~~! 更新: answers: 1. 120,quad.II 2. -40, quad.IV 3. 230, quad.III 4. sinΘ=3/5, cosΘ=4/5, tanΘ=3/4 5. sinΘ=-1/(sqrt2), cosΘ=1/sqrt2, tanΘ=-1 6. sinΘ=1/2, cosΘ=-(sqrt3)/2, tanΘ=-1/sqrt3 7. sinΘ=0, cosΘ=-1, tanΘ=0
最佳解答:
Quadrant I =0<θ<90 Quadrant II =90<θ<180 Quadrant III=180<θ<270 Quadrant IV =270<θ<360 sinθ=y/r, cosθ=x/r, tanθ=y/x [r=PO] So, 1. θ=90+30=120[Quadrant II] 2. θ=360-320=40[Quadrant I] 3. θ=270-40=230[Quadrant III] 4. sinθ=3/5, cosθ=4/5, tan=3/4 5. sinθ=-1/√2, cosθ=1/√2, tan=-1 6. sinθ=1/2, cosθ=-√3/2, tan=-1/√3 7. sinθ=0, cosθ=-1, tan=0 2011-08-19 21:02:16 補充： CORRECTION: 2. 320-θ=360, θ=-40[Quadrant IV] sorry for the mistake...
其他解答:
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