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三角函數廣義角的疑難!求解

Aug 25th 2014, 08:50

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tanθ = – 3/4 = 直線斜率考慮θ大小的兩種情況為 sinθ=3/5 或 sinθ = – 3/5 則sinθ/(1--tanθ) =(3/5)/(1–(–3/4))或sinθ/(1--tanθ) =–(3/5)/(1–(–3/4)) 得sinθ/(1--tanθ) =12 / 35或 sinθ/(1<sep/>...

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