# Quod Erat Demonstrandum

## 2012/03/16

### 好像是 cosine law

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

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

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

$a^2-c^2$

$=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$

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

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

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

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$D$ 點使 $\Delta DBC$ 為等邊三角形。

## 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 | 回應

• 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 | 回應

• 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 | 回應