# Quod Erat Demonstrandum

## 2009/02/19

### Just an old question about F.4 trigonometry

Filed under: Additional / Applied Mathematics,HKCEE — johnmayhk @ 9:55 下午
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$\sin1^o \times \sin2^o \times \sin3^o \times \dots \times \sin90^o$

Here is a way.

Knowing that

$\sin\theta \times \sin(60^o - \theta) \times \sin(60^o + \theta) \equiv \frac{1}{4}\sin3\theta$

Hence the required value

$= (\sin1^o\sin59^o\sin61^o)(\sin2^o\sin58^o\sin62^o)\dots(\sin29^o\sin31^o\sin89^o)(\sin30^o\sin30^o\sin90^o)\sin60^o$
$= (\frac{1}{4})^{30}\sqrt{3}(\sin3^o\sin6^o\sin9^o\dots\sin87^o)$
$= (\frac{1}{4})^{30}\sqrt{3}(\sin3^o\sin57^o\sin63^o)(\sin6^o\sin54^o\sin66^o)\dots(\sin27^o\sin33^o\sin87^o)\sin30^o\sin60^o$
$= (\frac{1}{4})^{40}\times 3(\sin9^o\sin18^o\sin27^o\dots\sin72^o\sin81^o)$
$= (\frac{1}{4})^{40}\times 3(\sin9^o\cos9^o)(\sin18^o\cos18^o)(\sin27^o\cos27^o)(\sin36^o\cos36^o)\sin45^o$
$= (\frac{1}{4})^{42}\times \frac{3\sqrt{2}}{2}\sin18^o\sin36^o\sin54^o\sin72^o$
$= (\frac{1}{4})^{42}\times \frac{3\sqrt{2}}{2}\sin36^o\cos18^o$
$= (\frac{1}{4})^{45}\times 6\sqrt{10}$

Of course, this is only one of the ways…

[OT] 今天我沒有上庭，太太轉述：原先法官判四至六個月，但一定要感謝法援的翁靜晶，她聽了我們的情況，竟然也為這個陌生人流淚，替我們求情，感恩地，法官先判兩星期羈留侯審。