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標題:
求中三級三角比問題解答
發問:
証明以下之三角恆等式﹕ tan2? -sin2? ≡ tan2? -sin2? 簡化以下問題﹕ (1+cos?θ)/sin^2?θ -1/(1-cos?θ )
最佳解答:
? = empty set φ = phi They are different symbols. 2014-09-28 20:57:44 補充： Please read: 圖片參考：https://s.yimg.com/rk/HA00430218/o/1265126181.png
其他解答:
(a+b)(a-b)=a^2 - b^2 sin^2 θ + cos^2 θ=1 (1+cos θ)/(sin^2 θ) - 1/(1-cos θ) =[ (1+cos θ)(1-cos θ) - (sin^2 θ) ]/ [ (sin^2 θ)(1-cos θ) ] =( 1 - cos^2 θ - sin^2 θ ) / [ (sin^2 θ)(1-cos θ) ] =( 1 - 1 ) / [ (sin^2 θ)(1-cos θ) ] =0/ [ (sin^2 θ)(1-cos θ) ] =0
求中三級三角比問題解答
發問:
証明以下之三角恆等式﹕ tan2? -sin2? ≡ tan2? -sin2? 簡化以下問題﹕ (1+cos?θ)/sin^2?θ -1/(1-cos?θ )
最佳解答:
? = empty set φ = phi They are different symbols. 2014-09-28 20:57:44 補充： Please read: 圖片參考：https://s.yimg.com/rk/HA00430218/o/1265126181.png
其他解答:
(a+b)(a-b)=a^2 - b^2 sin^2 θ + cos^2 θ=1 (1+cos θ)/(sin^2 θ) - 1/(1-cos θ) =[ (1+cos θ)(1-cos θ) - (sin^2 θ) ]/ [ (sin^2 θ)(1-cos θ) ] =( 1 - cos^2 θ - sin^2 θ ) / [ (sin^2 θ)(1-cos θ) ] =( 1 - 1 ) / [ (sin^2 θ)(1-cos θ) ] =0/ [ (sin^2 θ)(1-cos θ) ] =0
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