Quod Erat Demonstrandum



Filed under: Additional / Applied Mathematics,HKCEE — johnmayhk @ 4:01 下午

When giving the following basic trigonometry question in F.4 additional mathematics lesson:

Given A + B + C = 90^o, prove that \tan A\tan B + \tan B\tan C + \tan C\tan A = 1.

It should be extremely easy, just write A + B = 90^o - C \Rightarrow \tan(A + B) = \tan(90^o - C), however, a student gave

A + B + C = 90^o
2(A + B + C) = 180^o
2t = 180^o (where t = A + B + C)
\tan(2t) = \tan180^o
\frac{2\tan t}{1 - \tan^2t} = 0
\tan t = 0

Oops, t = 90^o actually, that means the equation above telling us

\tan90^o = 0

but, we all know that \tan90^o is undefined, so what’s wrong?

P.S. Patrick, I can’t update my blog too often now, because I’m busy working in a job and my family is moving to a new flat, so I can’t access internet at home recently.

5 則迴響 »

  1. 呀sir,條題目係 tanAtanB + tanBtanC + tanCtanA=1 呀

    迴響 由 Carmen — 2009/01/09 @ 8:00 下午 | 回覆

  2. 這與tan^2 t =/= 0這個限制有關嗎?

    by the way,你要take care呀,最近你也好像沒什麼精神…

    迴響 由 Patrick Wong — 2009/01/09 @ 11:56 下午 | 回覆

  3. Is there a typo in the original question? It seems to be \tan A \tan B + \tan B \tan C + \tan C \tan A = 1.

    迴響 由 Soarer — 2009/01/10 @ 12:06 上午 | 回覆

  4. Thank you Carmen and Soarer, I’d amended.

    Patrick, I’m extremely exhausted recently, thank you for your care.

    When it comes to the question, we may think about that, is

    \tan2t = \frac{2\tan t}{1 - \tan^2t}

    always true for any value of t?

    迴響 由 johnmayhk — 2009/01/10 @ 11:20 上午 | 回覆

  5. I just reference to the A.Math textbook.
    The compound angle formula of tangent is the quotient relation of the compound angle formulae of sine and cosine.
    And in one step, dividing cosAcosB to get the formula for tangent.
    The problem is here, cosAcosB must not be zero.
    In the case of tan(t+t), cos t = 0, as t=A+B+C=90 degree.

    迴響 由 C — 2009/01/12 @ 9:45 下午 | 回覆

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