Here’s a triangle:

I’ll use the vertex labels to also denote the corresponding angle measures.

Take the sines of all 3 angles in the triangle: , , and .

Would you believe it if I told you that these three sines satisfy the triangle inequality?

Even more, would you believe it if I told you that a triangle with sides of length , , and is similar to the original triangle?

Please think about this before reading further.

SPOILER ALERT!

Do you recognize the statement as implied by the law of sines?

In fact, the law of sines even tells us that the similarity ratio is equal to the diameter of the circle circumscribed about the original triangle.

We can use this understanding to prove the law of sines. First, note that in a circle of diameter 1, an inscribed angle measuring degrees subtends a chord of length (see figure at left). To see this, note that the blue chord has length because the central angle of the arc subtended by the inscribed angle with measure degrees has measure degrees.

That means that if we scale a triangle so that its circumscribed circle has unit diameter, the lengths of the 3 sides of the scaled triangle must be the sines of its 3 angles, and since we know what the corresponding sides are, we get the law of sines, complete with similarity ratio being the diameter of the circumscribed circle.

### The Law of Cosines

Just for kicks, we can use the law of sines to give the following proof of the law of cosines.

In a triangle, the angles add up to 180 degrees, so . Using this, we can compute using standard trigonometric identities:

And that’s the law of cosines. (Because the equation is homogeneous in the sines, the law of sines informs us that we can replace with , with , and with , to get an equivalent equation, which is how the law of cosines is usually written.)

In general, if you have an equation that is homogeneous in the lengths of the sides of a triangle, you automatically have an equation in terms of the sines of the angles of the triangle, via the law of sines. For example, by dropping the altitude from vertex in the yellow triangle, we see that . The lengths all appear with exponent 1, so, via the law of sines, we can replace with , with , and with to get , and since , we also have , and, voila!, out pops the trigonometric identity for the sine of a sum. (Here, we needed angles and to be positive and have a sum less than 180 degrees, so one has to think a little more to get the trig identity without restriction.)

### Ptolemy’s Theorem

Ptolemy’s theorem relates the lengths of the sides of a cyclic quadrilateral and the lengths of its diagonals. The equation is homogeneous in these lengths, so we can immediately convert to a trigonometric identity. We label the vertices and angles as in the red circle. Notice that because they subtend the same arc. So the 4 unlabeled angles in the corners are equal to the labeled ones in some order.

Ptolemy’s theorem says that . Using the law of sines (which can be envisioned as scaling the figure so that the circle has diameter 1), this identity involving lengths is converted into a trigonometric identity:

.

Turning this around, Ptolemy’s theorem can be proven by proving this trig identity, and to prove it, one only needs a few basic trig identities and the fact that . Try it and see for yourself!

Now, check out problem #15 from the 2013 AIME II.