# Given $\sin{A}:\sin{B}:\sin{C}$ how to find $\sin{(2A)}:\sin{(2B)}:\sin{(2C)}$?

Question:

Given $x,y,z>0$, and the positive number $k$ such $$\dfrac{x^2}{x^2+k}+\dfrac{y^2}{y^2+k}+\dfrac{z^2}{z^2+k}=1$$ in $\Delta ABC$,
$$(1):\sin{A}:\sin{B}:\sin{C}=\dfrac{x}{x^2+k}:\dfrac{y}{y^2+k}:\dfrac{z}{z^2+k}$$ $$(2): \sin{(2A)}:\sin{(2B)}:\sin{(2C)}=\dfrac{1}{x^2+k}:\dfrac{1}{y^2+k}:\dfrac{1}{z^2+k}$$ show that: $$(1)\Longrightarrow (2)$$

this problem is from I write a paper, and this result seems interesting.

My idea:

and if $x=3,y=4,z=18$,then $$\dfrac{9}{9+k}+\dfrac{16}{16+k}+\dfrac{324}{324+k}=1,k>0$$

then $$k=96$$see:equation

then $$\sin{A}:\sin{B}:\sin{C}=4:5:6$$ see:2 so use sine therom $$a:b:c=4:5:6$$ let $$a=4t,b=5t,c=6t$$ so $$\cos{A}=\dfrac{b^2+c^2-a^2}{2bc}=\dfrac{3}{4},\cos{B}=\dfrac{a^2+c^2-b^2}{2ac}=\dfrac{9}{16},\cos{C}=\dfrac{a^2+b^2-c^2}{2ac}=\dfrac{1}{8}$$

so $$\cos{A}:\cos{B}:\cos{C}=12:9:2$$ so $$\sin{2A}:\sin{2B}:\sin{2C}=(4\cdot 12):(5\cdot 9):(6\cdot 2)=16:15:4$$ and other hand $$\dfrac{1}{x^2+k}:\dfrac{1}{y^2+k}:\dfrac{1}{z^2+k}=\dfrac{1}{9+96}:\dfrac{1}{16+96}:\dfrac{1}{324+96}=16:15:4$$ see 3

second example:

let $x=\sqrt{2},y=\sqrt{5},z=\sqrt{10}$,then we easy to find $k=10$ see 4, and $$\sin{A}:\sin{B}:\sin{C}=\dfrac{\sqrt{2}}{12}:\dfrac{\sqrt{5}}{15}:\dfrac{\sqrt{10}}{20}=5\sqrt{2}:4\sqrt{5}:3\sqrt{10}$$ so $$\cos{B}=\dfrac{a^2+c^2-b^2}{2ac}=\dfrac{50+90-80}{2\times 5\sqrt{2}\times 3\sqrt{10}}=\dfrac{1}{\sqrt{5}}$$ $$\cos{A}=\dfrac{1}{\sqrt{2}},\cos{C}=\dfrac{1}{\sqrt{10}}$$ so $$\sin{2A}:\sin{2B}:\sin{2C}=5:4:3$$ other hand $$\sin{2A}:\sin{2B}:\sin{2C}=\dfrac{1}{x^2+k}:\dfrac{1}{y^2+k}:\dfrac{1}{z^2+k}=\dfrac{1}{12}:\dfrac{1}{15} :\dfrac{1}{20}=5:4:3$$ so this case is also true

But I can't prove this.can you help?

• Your title is not clear: what is a konwsin (or perhaps konw)? – Rory Daulton Oct 26 '14 at 11:24

Let $$p = \frac{x^2}{x^2 + k}, \quad q = \frac{y^2}{y^2 + k}, \quad r = \frac{z^2}{z^2 + k}.$$
Then by assumption $p + q + r = 1$.
Letting $a$, $b$ and $c$ be the sides of the triangle, we have by the sine law \begin{align*} a^2 : b^2 : c^2 &= \sin^2 A : \sin^2 B : \sin^2 C \\ &= \frac{x^2}{(x^2 + k)^2} : \frac{y^2}{(y^2 + k)^2} : \frac{z^2}{(z^2 + k)^2} \\ &= \frac{x^2}{x^2 + k}\frac{k}{x^2 + k} : \frac{y^2}{y^2 + k}\frac{k}{y^2 + k} : \frac{z^2}{z^2 + k}\frac{k}{z^2 + k} \\ &= p(1-p):q(1-q):r(1-r) \\ &= p(q + r):q(p + r): r(p + q). \end{align*}
Now by the sine and cosine laws, we have (using the symbol $\propto$ to mean "is proportional to"), \begin{align*} \sin 2A &= 2\sin A \cos A \\ &\propto a\frac{b^2 + c^2 - a^2}{2bc} \\ &\propto a^2(b^2 + c^2 - a^2) \\ &\propto p(q+r)[q(p+r) + r(p+q) - p(q + r)] \\ &=2pqr(q+r) \\ &\propto q + r \\ &= 1 - p \\ &= \frac{k}{x^2 + k} \\ &\propto \frac{1}{x^2 + k}. \end{align*}