數學Trigonometric Relations

2009-05-03 7:56 am
tan 70=(1+0.8 sinθ) / 0.8 cosθ

回答 (1)

2009-05-03 9:22 am
✔ 最佳答案

tan70o = (1 + 0.8 sinθ) / 0.8cosθ

0.8tan70ocosθ = 1 + 0.8sinθ

0.8tan 70o[cos2(θ/2) - sin2(θ/2)] = [cos2(θ/2) + sin2(θ/2)] + 0.8[2sin(θ/2)cos(θ/2)]

(1 + 0.8tan70o)sin2(θ/2) + 1.6sin(θ/2)cos(θ/2) + (1 - 0.8tan70o)cos2θ = 0

[(1 + 0.8tan70o)sin2(θ/2) + 1.6sin(θ/2)cos(θ/2) + (1 - 0.8tan70o)cos2(θ/2)]/cos2(θ/2) = 0

(1 + 0.8tan70o)tan2(θ/2) + 1.6tan(θ/2) + (1 - 0.8tan70o) = 0

tan(θ/2) = {-1.6 √[(1.6)2 - 4(1 + 0.8tan70o)(1 - 0.8tan70o)]}/[2(1 + 0.8tan70o)]

tan(θ/2) = 0.4110 or tan(θ/2) = -0.9114

θ/2 = 22.34o or θ/2 = 180o - 42.34o

θ = 44.68o or θ = 275.32o
=


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