Quod Erat Demonstrandum

2012/03/16

好像是 cosine law

Filed under: Additional / Applied Mathematics,NSS — johnmayhk @ 5:05 上午

中學同學熟識 Cosine law:

a^2=b^2+c^2-2bc\cos A

等。

之前看過,也頗有「美感」的 law:

a^2+b^2-2ab\cos(C+\frac{\pi}{3})=b^2+c^2-2bc\cos(A+\frac{\pi}{3})=c^2+a^2-2ca\cos(B+\frac{\pi}{3})

同學,先試試證明吧。

暫時想不到擬題者出題之原理,唯用比較醜陋的方法做:

a^2-c^2

=a^2-c^2-a^2\sin^2 C+c^2\sin^2 A\because a\sin C=c\sin A

=a^2\cos^2 C-c^2\cos^2 A

=(a\cos C+c\cos A)(a\cos C-c\cos A)

=b(a\cos C-c\cos A)\because b=a\cos C+c\cos A

=b(a\cos C-c\cos A-a\sqrt{3}\sin C+c\sqrt{3}\sin A)\because a\sin C=c\sin A

=b(a(\cos C-\sqrt{3}\sin C)-c(\cos A-\sqrt{3}\sin A))

=2ba\cos(C+\frac{\pi}{3})-2bc\cos(A+\frac{\pi}{3})

a^2+b^2-2ab\cos(C+\frac{\pi}{3})=b^2+c^2-2bc\cos(A+\frac{\pi}{3})

**************************************
後記:以下是網友 Mt 提供的「無言證明」,相信才是出題者之原意。

參下圖:

D 點使 \Delta DBC 為等邊三角形。

分別在 \Delta ADB\Delta ADC,以 cosine law 考慮線段 AD 的長度,立即得出結果,完全不用醜陋的甚麼 compound angle formula 和代數運算,感謝!

4 則迴響 »

  1. 1. Contruct an isosceles trianlge with the base BC.
    2. Let the vertex of this triangle be D.
    3. CD = a
    AC^2 + CD^2 – 2(AC)(CD)cos(C+Pi/3) = AD^2

    => b^2 + a^2 -2ab*cos(C+Pi/3)= AD^2

    Apply the cosine law to the triangle ABD, we have
    c^2 + a^2 – 2ac*Cos(B+Pi/3)= AD^2

    迴響 由 Mt — 2012/03/17 @ 3:48 上午 | 回覆

    • Thank you Mt!!

      But when refer to the following case,

      why I got

      a^2+b^2-2ab\cos(\frac{\pi}{3}-C)=c^2+a^2-2ca\cos(\frac{\pi}{3}-B)

      ?

      迴響 由 johnmayhk — 2012/03/17 @ 10:30 上午 | 回覆

      • With respect to the line BC, D and A should be on the opposite side.

        迴響 由 Mt — 2012/03/17 @ 11:13 上午

  2. Thanks!

    迴響 由 johnmayhk — 2012/03/17 @ 4:08 下午 | 回覆


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