航空百科

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標題:

sin cos tan

發問:

1)tanθ/sinθcosθ 2)-5sin2θ-5cos2θ 3)sinθ=cos35° 求解筨 thx

最佳解答:

1) tanθ/sinθcosθ =(sinθ/cosθ)/sinθcosθ =sinθ/cosθ * 1/sinθcosθ =1/cos2θ 2) - 5sin2θ - 5cos2θ = - 5(1 - cos2θ) - 5cos2θ = - 5 + 5cos2θ - 5cos2θ = -5 3) sinθ=cos 35° θ=55° or θ=125° 2015-01-01 21:11:34 補充： Or you can answer question 3 like this, sinθ=cos 35° sinθ=sin(90° - 35°) sinθ=sin 55° θ=55° or θ=180° - 55° = 125° (∵sin 55° = sin 125°)

其他解答:6CC7293C79127CE5