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菜圃 asked in 社會與文化語言 · 1 decade ago

the angle of intersection

三角函數問題, 希望高手幫忙解題

Find the acute angle of intersection of the curves y= sin 2x and y = cos x at point (Π/2, 0)

謝謝!

Update:

根據 tan 的和角公式得知,兩條切線的夾角之 tan 值會是

(m2 - m1)/(1 + m1*m2) = 1 / 3

因此兩條曲線的交角為 arctan (1/3) 大約是 18.4349 度

為什麼和角公式回事兩條切線的夾角呢??

1 Answer

Rating
  • linch
    Lv 7
    1 decade ago
    Favorite Answer

    找兩條曲線在 (π/2, 0)這點的交角

    先求切線斜率

    y = sin 2x ==> y' = 2 cos 2x ==> 在(π/2, 0) y' = - 2 = m1

    y = cos x ==> y' = - sin x ==> 在(π/2, 0) y' = - 1 = m2

    根據 tan 的和角公式得知,兩條切線的夾角之 tan 值會是

    (m2 - m1)/(1 + m1*m2) = 1 / 3

    因此兩條曲線的交角為 arctan (1/3) 大約是 18.4349 度

    2009-03-04 19:55:05 補充:

    若一條直線與正向 x 軸逆時針夾角為 A 則斜率即為 tan A

    所以你現在有兩條切線一條與正向 x 軸逆時針夾角為 A,另一條與正向 x 軸逆時針夾角為 B ( 夾角為 B - A )

    則 tan A = m1, tan B = m2

    又由 tan(B - A) = (m2 - m1)/(1 + m1*m2)

    所以可得 B - A 之 tan 值

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