<s>[INST] Parallelogram trigonometry (Sorry for the ambiguous title, couldn't think of a better one)
While leafing through a highschool textbook, I found what looked like an interesting question in trigonometry. My trigonometry skills are borderline 0, but I didn't expect it to be too much of a challenge. Well, I was wrong:
The sides of a parallelogram are $a$ and $b$ and its sharp angle is $\alpha$. The diagnols are $n$ and $m$, and the sharp angle between them is $\beta$.
A. Prove: $\frac{mn}{2ab} = \frac{\sin\alpha}{\sin\beta}$
B. Let: $\alpha = \beta$, $a < b$, $m < n$
Prove: $6a^2 + 2b^2 = 3m^2+n^2$
And in (rough) drawing:

Following the law of cosines (and that $\cos(180-\theta) = -\cos(\theta)$):
$n^2 = a^2 + b^2 - 2ab \cos\alpha$ (in $\Delta ABC$)
$m^2 = a^2 + b^2 - 2ab \cos(180-\alpha) = a^2 + b^2 + 2ab \cos(\alpha)$ (in $\Delta DAC$)
$a^2 = (\frac{m}{2})^2 + (\frac{n}{2})^2 -2 \frac{m}{2} \frac{n}{2} \cos(\beta)$ (in $\Delta AEB$)
$b^2 = (\frac{m}{2})^2 + (\frac{n}{2})^2 -2 \frac{m}{2} \frac{n}{2} \cos(180 - \beta)$ (in $\Delta BEC$)
Expanding the last two equations:
$$a^2 = \frac{m^2}{4} + \frac{n^2}{4} - \frac{mn \cos(\beta)}{2}$$
$$b^2 = \frac{m^2}{4} + \frac{n^2}{4} + \frac{mn \cos(\beta)}{2}$$
$$\Rightarrow$$
$$a^2 + b^2 = \frac{m^2}{2} + \frac{n^2}{2}$$
And that's where I hit a wall. I have six variables, and can't find a way to express them in a fashion which resembles the end result. A major setback is that I couldn't find a way to express both alpha and beta in the same triangle - if I could, then the law of sines will probably be a rescuer.
If possible, I'd like that instead of solving it, maybe you can show me a guideline - where I went wrong, or what I'm missing. Thank you in advance.
 [/INST] For part A, try counting the area of the parallelogram in two different ways, as suggested by Jim Belk.
For part B, notice that your diagram has $n$ and $m$ reversed, since $m<n$. In particular, $\alpha$ should be opposite $m$, not $n$. The modified version of your formula for $m$ is then
$$m^2 = a^2+b^2 -2ab\cos\alpha.$$
Try combining this with your formula
$$4a^2 = m^2 + n^2 -2mn\cos\beta$$
and use the result from part A.
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