# Why is $\frac{k \cdot(\sin (A) + \sin(B))}{k \cdot(\sin(A) -\sin(B))} = \frac{a+b}{a-b}$ valid in a triangle?

This is extracted from my module :

In a $\displaystyle\bigtriangleup ABC$, $$\displaystyle \frac{\sin (A) + \sin(B)}{\sin(A) -\sin(B)} = \frac{k \cdot(\sin (A) + \sin(B))}{k \cdot(\sin(A) -\sin(B))} = \frac{a+b}{a-b}$$

where $A,B,C$ are the angles and $a,b,c$ are the opposite sides,I can't figure out how this step holds valid?

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It's the sine rule; also $\sin(a)$ should be $\sin)A)$. – Robin Chapman Nov 16 '10 at 10:50
What is k supposed to be, and why are you taking the sine of a side length? – J. M. Nov 16 '10 at 10:52
@ J. M.:I got it from Robin's comment $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k$(say) then applying the rule of propotion ... – Quixotic Nov 16 '10 at 10:57
Anyway, the derivation of the law of tangents here should be useful. – J. M. Nov 16 '10 at 10:57

From the $\sin$ rule, you know that

$$\frac{a}{\sin A}=\frac{b}{\sin B}.$$

In a proportion such as

$$\frac{p}{q}=\frac{r}{s},$$

we have (by a property of proportions, which you can check algebraically):

$$\frac{p+r}{p-r}=\frac{q+s}{q-s}.$$

Setting $p=a$, $q=\sin A$, $r=b$, $s=\sin B$ gives you

$$\frac{a+b}{a-b}=\frac{\sin A+\sin B}{\sin A-\sin B}.$$

$$\dfrac{p+r}{p-r}=\dfrac{q+s}{q-s}\Leftrightarrow \left( p+r\right) \left( q-s\right) -\left( p-r\right)\left( q+s\right) =0$$

$$\Leftrightarrow 2qr-2ps=0\Leftrightarrow qr=ps\Leftrightarrow \dfrac{p}{q}=\dfrac{r}{s}.$$

And clearly, for $k\neq 0$

$$\dfrac{q+s}{q-s}=\dfrac{k\left( q+s\right) }{k\left( q-s\right) }.$$

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I'm not sure what $k$ is supposed to be, but another way to see that the first and third expressions are equal is to rewrite the first one as $$\frac{\frac{\sin(A)}{\sin(B)}+1}{\frac{\sin(A)}{\sin(B)}-1}$$ and then note that the law of sines tells you $$\frac{\sin(A)}{\sin(B)} = \frac{a}{b}$$.

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The sine rule for a triangle goes as

$\frac{a}{\sin(A)} = \frac{b}{\sin(B)} = \frac{c}{\sin(C)} = 2R$, where $R$ is the radius of the circumcircle.

From this, we get $\sin(A) = \frac{a}{2R}$ and $\sin(B) = \frac{b}{2R}$

The problem has $\frac{\sin(A) + \sin(B)}{\sin(A) - \sin(B)}$. Plug in the expressions for $\sin(A)$ and $\sin(B)$ from the above and we get,

$\frac{\frac{a}{2R}+\frac{b}{2R}}{\frac{a}{2R}-\frac{b}{2R}} = \frac{\frac{a+b}{2R}}{\frac{a-b}{2R}} = \frac{a+b}{a-b}$

Hence, $\frac{\sin(A) + \sin(B)}{\sin(A) - \sin(B)} = \frac{a+b}{a-b}$

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Geometrically, this is not too hard to see. Presumably, you understand why

$$\frac{\sin(A) + \sin(B)}{\sin(A) - \sin(B)} = \frac{k(\sin(A) + \sin(B))}{k(\sin(A) - \sin(B))}$$

Let k be the length of the normal from C onto AB. What is then k sin(A) and k sin(B)?

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