Mathematical Tripos Examination for 1878

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สำหรับข้อ 8 เฉลยยาวมากๆ ขอต๊ะไว้ก่อน ว่างๆ จะมาโพสต์ให้อ่านกัน

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มาแล้วครับ ... เฉลยข้อ $8$ พิมพ์ยากน่าดู


เฉลยข้อสอบชุดที่ 2: MONDAY, December 31, 1877. 13:30 to 16:00

8. Prove geometrically
$\;\;\; (i) \;\;\; \sin A+\sin B = 2\sin\frac12(A+B)\cos\frac12(A-B),$
$\;\;\; (ii) \;\; \cos A-\cos B = 2\sin\frac12(B+A)\sin\frac12(B-A);$
and explain how such formulae are shewn to be true universally for all magnitudes of the angles $A$ and $B$.
If $$\;\;\; \frac{\cos(\alpha+\beta+\theta)}{\sin(\alpha+\beta)\cos^2\gamma} = \frac{\cos(\gamma+\alpha+\theta)}{\sin(\gamma+\alpha)\cos^2\beta},$$
and $\beta$ and $\gamma$ are unequal, prove that each member will be equal to
$$\frac{\cos(\beta+\gamma+\theta)}{\sin(\beta+\gamma)\cos^2\alpha},$$
and that
$$ \cot\theta = \frac{\sin(\beta+\gamma)\sin(\gamma+\alpha)\sin(\alpha+\beta)}{\cos(\beta+\gamma)\cos(\gamma+\alpha)\cos(\alpha+\beta) + \sin^2 (\alpha + \beta + \gamma)}.$$

Solution:

$\;\;\;$ From the given equation
$$ \cot\theta = \frac{\sin(\alpha+\beta)\sin(\gamma+\alpha)\cos^2\beta - \sin(\gamma+\alpha)\sin(\alpha+\beta)\cos^2\gamma}{\cos(\alpha+\beta)\sin(\gamma+\alpha)\cos^2\beta - \cos(\gamma+\alpha)\sin(\alpha+\beta)\cos^2\gamma}.$$

of which the numerator = $\sin(\beta+\gamma)\sin(\gamma+\alpha)\sin(\alpha+\beta)\sin(\gamma-\beta)$,

and the denominator

$\;\;\; = \frac12\{\cos(\alpha+\beta)\sin(\gamma+\alpha) - \cos(\gamma+\alpha)\sin(\alpha+\beta)$
$\;\;\;\;\;\;\;\;\;\;\;\; + \cos(\alpha+\beta)\sin(\gamma+\alpha)\cos2\beta - \cos(\gamma+\alpha)\sin(\alpha+\beta)\cos2\beta \}$

$\;\;\; = \frac14\{2\sin(\gamma-\beta) + \sin(2\alpha+\beta+\gamma)(\cos2\beta-2\cos2\gamma) + \sin(\gamma-\beta)(\cos2\beta+\cos2\gamma) \}$

$\;\;\; = \frac14\sin(\gamma-\beta)\{2 + \cos2\alpha + \cos2\beta + \cos2\gamma) - \cos(2\alpha+2\beta+2\gamma) \}$

$\;\;\; = \frac14\sin(\gamma-\beta)\{4\sin^2(\alpha+\beta+\gamma) + \cos2\alpha + \cos2\beta + \cos2\gamma) + \cos(2\alpha+2\beta+2\gamma) \}$

$\;\;\; = \sin(\gamma-\beta)\{\sin^2(\alpha+\beta+\gamma) + \cos(\alpha+\beta)\cos(\beta+\gamma)\cos(\gamma+\alpha) \}$

Throwing out the factor $\sin(\gamma-\beta)$, which is common to the numerator and denominator, this becomes the expression for $\cot\theta$ in the question, and since it involves $\alpha,\;\beta,\;\gamma$ symmetrically it follows that each side of the given equation also
$$ = \frac{\cos(\beta+\gamma+\theta)}{\sin(\beta+\gamma)\cos^2\alpha}.$$

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เฮ้อ...กว่าจะพิมพ์ข้อ 8 เสร็จตาลายน่าดู
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