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Besides one of those laws: Like there's a proof for the law of cosines that gives a hint towards a proof for law of sines. Besides that. But something like that.

To clarify no calculus, non-euclidean geometry or linear algebra.

I guess vectors and matrix algebra (that's linear algebra without the proofs?) Would be okay but try to keep it simple.

Despite having a master's degree I don't know what keywords to use. :(

BCLC
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    Not sure if this is what you have in mind but perhaps a derivation of a formula for the area of a triangle. Suppose a triangle has sides $a$ and $b$ and the angle between them is $\theta$, find the area of the triangle. Maybe something in 3D. Is this just for a proof exercise or do you want to prove something that is useful in other Geometry problems? – Addem Dec 09 '16 at 15:27
  • @Addem Applying law of cosines to derive area of triangle? If so, no. If not, what do you mean? I'm looking for something we can prove based on a proof of law of sines or a proof of law of cosines. For example, I can't seem to find it right now, but I saw a proof of law of sines based on proof of law of cosines in acute case. To clarify, what I mean consider... – BCLC Dec 22 '16 at 16:30
  • @Addem ... this exercise requiring a proof of a proposition on infinite products presented after proof of BCL2. BCL2 doesn't seem to help (well one user tried, but answer is now deleted), but my proof is based on proof of BCL2. – BCLC Dec 22 '16 at 16:31

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Suppose they know that the usual metric in $\mathbb{R}^2$ is given by:

$$d((x_0,y_0),(x_1,y_1)) = \sqrt{(x_1-x_0)^2+(y_1-y_0)^2}$$

and given a circle of radius $r$, it's points are of the form $(r \cos \theta, r \sin \theta)$. Then consider the triangle determined by $A = (0,0), B = (a,0)$ and $C$ which is of length say $b$ (located to the left of $A$). Hence, $C$ lies on the circle of radius $b$. Let $\theta$ be the angle made at the segments $\bar{AC}$ and $\bar{AB}$. Then $C = (b \cos \theta, b \sin \theta)$ and so we have:

$$d\big((b \cos \theta, b \sin \theta),(a,0)\big) = \sqrt{(a-b \cos \theta)^2+ b^2 \sin^2 \theta} = \sqrt{a^2+b^2 - 2ab \cos\theta} $$

$$\\$$

enter image description here

Hence, if we set $\textbf{length}(\bar{BC}) = c$ (given above) then $c^2 = a^2+b^2 - 2ab \cos\theta$ and this is the result for Law of Cosines. You can play this game now with other sides be unknown i.e the proof only requires the two facts above.

Faraad Armwood
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