Quod Erat Demonstrandum


Trivial sharing on trigonometry

Filed under: Additional / Applied Mathematics,mathematics,NSS — johnmayhk @ 7:33 上午

What’s wrong with the following steps?

\sin\theta < 0

\Rightarrow \theta lies in the 3rd or 4th quadrant

\Rightarrow 180^o < \theta < 360^o

\Rightarrow 90^o < \frac{\theta}{2} < 180^o

\Rightarrow \sin\frac{\theta}{2} > 0

For questions like: “If \sin\theta = -\frac{1}{2}, evaluate \sin\frac{\theta}{2}“, we need to consider the sign of \sin\frac{\theta}{2}.

As you may see that the incorrect steps above occur at

\theta lies in the 3rd or 4th quadrant

\Rightarrow 180^o < \theta < 360^o

A correct expression is suggested as

\theta lies in the 3rd or 4th quadrant

\Rightarrow \theta = 360^on + \alpha (for some integer n and 180^o < \alpha < 360^o)

Hence \sin\frac{\theta}{2} > 0 may NOT be always true.

Urm… I’d made similar mistake too, the sharing is a reminder to me… ^_^"


4 則迴響 »

  1. That’s the mistake i made in the test, i think i hadn’t thought about the possibility of sinX/2 would be negative in the test.

    迴響 由 Matthew — 2010/05/13 @ 8:05 下午 | 回應

    • Not only you, some students and even I had made the mistake too! And the mistake made you lost 1 mark, otherwise you’ll obtain full mark in the quiz! Good job Matthew!!

      迴響 由 johnmayhk — 2010/05/13 @ 10:49 下午 | 回應

  2. thanks

    迴響 由 Matthew — 2010/05/13 @ 11:02 下午 | 回應

  3. Just a little bit following up.

    Perhaps we should be careful in proving that \sin x and \tan \frac{x}{2} are of the same sign for any real number x.

    迴響 由 johnmayhk — 2010/05/18 @ 2:21 下午 | 回應

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