Making sense of the law of sines
I was recently thinking about an old equation — the law of sines — when I stumbled upon an elegant perspective that I’d never seen before. This new point of view adds a stronger intuition for why the law is true, and it generalizes the law to other shapes — not just triangles.
To begin, let’s take a look at the original law of sines.
Property 1 [Original law of sines]
Suppose a triangle has side lengths a, b, and c; and each of these sides is opposite the angles ∠A, ∠B, and ∠C.
Then
The property is easier to remember if you think of it as
where a constant is some value that stays the same for whichever angle you plug into the left side — although this constant is not the same for different triangles!
Example
Consider a 45º–90º–45º triangle with sides a and b of length 1, and hypotenuse c of length √2:
For this triangle, the constant turns out to be 1/√2. If you’re like me, you’re probably wondering, “Hey! What does that value mean? Is there some intuition behind the magic constant in equation (∗) above?”
Sometimes a proof sheds more light on what’s going on behind the scenes of an equation — so let’s take a look at a traditional proof of the law of sines.
The proof I have in mind depends on an area formula that may be new to you, so I’ll provide it first.
Property 2 [The area of a triangle]
The area of a triangle with side lengths a and b, and angle ∠C between them, is given by
The proof of this area formula is tangential to the main line of thought, so I’ll skip it for now. It’s included at the end of the post if you’re curious.
This area formula is enough to prove the law of sines.
Proof of property 1 [Original law of sines]
Suppose a triangle has side lengths a, b, and c; and each of these sides is opposite the angles ∠A, ∠B, and ∠C.
The area can be expressed in three different ways using property 2:
Divide through by abc to arrive at
Aha! Based on this proof, the magic constant from the law of sines must have the intuitive meaning of:
Wait. What? I have no idea what the value abc means conceptually. This proof has not helped my intuition at all!
Ok, let’s try another approach. The next property is the keystone of this post.
I’ll use the notation ‖xy‖ to indicate the distance between x and y.
Property 3
Suppose the points x, y, and z lie on a circle of radius r. Then,
In other words, if we flip the ratios in the traditional law of sines, we arrive at
The value 2r is the diameter of the circle that hits all three corners of the triangle — a nice, meaningful quantity! This reveals that the previous version of the law of sines had a ratio of 1/2r, a fraction begging to be inverted.
Besides offering this new way to prove the law of sines, property 3 also offers a cool generalization that I’ll explore in a moment. Before that, I want to prove property 3 itself. The proof relies on a simple relationship between angles inside a circle called the Central Angle Theorem.
Up until now, I’ve used the notation ∠A with a single variable A to indicate an angle. This notation is short, but can be ambiguous if there are multiple angles that may be indicated by the single variable. For added clarity, I’ll sometimes use the notation ∠xzy to refer to the angle at z between the lines xz and zy.
Property 4 [Central Angle Theorem]
If points x, y, and z lie on a circle with center c, then angles ∠xcy and ∠xzy can be expressed in terms of each other as in the following two cases.
Case 1: z is not between x and y; then ∠xcy = 2∠xzy.
Case 2: z is between x and y; then∠xcy = 2(180º–∠xzy).
This is a great geometric fact to play with. Intuitively, it means the angle at the circle’s center completely determines the angle along the circle’s circumference, even if z has some freedom to move along the circle.
The proof of the Central Angle Theorem is not critical to the main line of thought, so I’ll defer it to the end of this post.
This is enough background to prove property 3, the equation ‖xy‖/sin(∠z)=2r.
Proof of property 3
This figure will help illustrate the reasoning:
For now, suppose that z is not between x and y. Use the letter θ to denote ∠z. Property 4 — the Central Angle Theorem — indicates that ∠xcy = 2θ. Let w be the point halfway between x and y. Notice that △xcy is isosceles since both lines xc and yc are radii of the circle. By symmetry, the line wc must cut △xcy into two congruent smaller triangles. In particular, ∠xcw = ∠wcy, which means both angles must be θ as their sum is 2θ. This symmetry also means line cw meets line xy at a right angle.
Let r be the radius of the circle. By the definition of sine applied to △xcw, sin(θ) = ‖xw‖/‖xc‖ = ‖xw‖/r. Thus ‖xy‖ = 2‖xw‖ = 2r sin(θ). Since θ=∠z, we can restate this as ‖xy‖/sin(∠z)=2r. This completes the proof when z is not between x and y.
What if z is between x and y? In this case, we can reuse most of the above proof, with one key difference being to define θ as 180º-∠z so that the appropriate case of the Central Angle Theorem results in the same angles for △xcw. Since triangles △xcw and △xcy are effectively the same as in the last case, the remainder of the proof is also effectively the same, keeping in mind that sin(∠z) = sin(180º-∠z) = sin(θ), using the fact that sin(α) = sin(180º-α) for any angle α.
This completes the proof that ‖xy‖/sin(∠z)=2r.
Now that I’ve proven property 3, I feel free to reap the rewards it has to offer. I noted above that it reveals a nice meaning for the law of sines. In fact, it also provides a generalized form that works for many other shapes. In particular, it applies to any cyclic polygon; a cyclic polygon is defined as a polygon whose corners all lie on the same circle.
Property 5 [Generalized law of sines]
Suppose a polygon has all its corners on a circle of radius r. If x, y, and z are any three corners of the polygon, then
This property doesn’t require a separate proof as it’s essentially a restatement of property 3. The only difference is that the wording emphasizes that any triangle △xyz in a circle of radius r obeys ‖xy‖/sin(∠xzy)=2r — and any three corners of a cyclic polygon form such a triangle with the same value of r.
Example
Consider the cyclic pentagon in this figure:
Applying property 5 to each of the sides in turn results in a law-of-sines-like series of equal expressions:
This clarifies how property 5 generalizes the original law of sines. The original law is a simultaneous equation involving the 3 sides of a triangle; property 5 gives a simultaneous equation between all the sides of a cyclic polygon, however many there are.
What other shapes does property 5 apply to? Well, every rectangle is a cyclic polygon, as is every regular polygon; as a reminder, a regular polygon has all its side lengths, and all its adjacent angles equal to each other. Equilateral triangles and squares are examples of regular polygons.
I’d like to define a circumradius for the next example; this is the radius of the circle which goes through all the corner points of a cyclic polygon. It’s the same value r referred to in property 5.
Another example
Property 5 can help you use a side length s to find the circumradius r of a regular n-gon, which is a regular polygon with n sides. Choose any two adjacent corners x and y. They form the angle θ=360º/n with the center point:
Let z be any other corner of the n-gon. Then ∠xzy = θ/2 = 180º/n by the central angle theorem. Then r = ‖xy‖/(2 sin(∠xzy)) = s/(2 sin(180º/n)).
Remember the original proof of the law of sines? The expression abc showed up — the product of all three side lengths of a triangle — and it appeared as a stumbling block to intuition. Now that things make more sense in light of property 3, it may be worth revisiting that expression.
Just after the proof of property 1, I had arrived at this equation:
Property 3 indicates that
Combining these, we get
This proves a new equation interesting enough to be called its own property.
Property 6 [New area of a triangle]
Suppose a triangle has side lengths a, b, c, and circumradius r. Then its area is given by
If the corners of a triangle are on a circle with radius 1 — called a unit circle— then this formula takes on the nice form
This is such a surprisingly simple expression that I questioned its truth and wanted to test it out. If the proofs are correct, then a test is not really necessary — but without the reward of a successful test, it feels like we did all the proof work for nothing!
The animation below shows a triangle moving within a unit circle. Although the two quantities for the area are always the same, the top value is computed as abc/4 while the bottom is deduced using a more well-known area equation called Heron’s formula. The fact that the two values always match is empirical evidence that the new area formula is indeed correct.
Several of the properties discussed in this post were new to me, although they are not new to the world; for example, here is one previous page which mentions them. In general, if a theorem has a short proof based on widely-known concepts, then it’s a safe bet that someone has already thought of it.
Nonetheless, a previously memorized-but-not-deeply-understood equation — the law of sines — has gained a new meaning for me, and I hope these thoughts may be interesting to anyone else who enjoys improving their own intuition around delicious mathematical morsels like these.
Appendix
In case you’re curious, I wrote a companion post on how to write a math post on Medium. Because Medium isn’t designed with mathematical layout in mind, writing this post was not easy. In the insanely unlikely case that you actually work at Medium, I also wrote a note on three features that would make Medium a better place for technical writers.
Now for the two proofs that were skipped earlier.
Proof of property 2 [Area of a triangle]
As a reminder, this is a proof that the area of a triangle with side lengths a and b, and angle C between those sides, can be expressed as ½ ab sin(∠C).
Consider the triangle here:
A standard triangle area formula is
Suppose the value of side lengths a and b, and angle C are known, but not the value of the height h. Then the definition of sine indicates that sin(∠C) = h/a. Thus, solving for h,
This completes the proof.
Proof of property 4 [Central Angle Theorem]
This proof uses 3 diagrams based on the placement of z relative to x and y.
Diagrams 1A and 1B cover the possibility that z is not between x and y; these correspond to case 1 in the statement property 4. These diagrams must be separate since △xzc may or may not overlap △yzc, and both cases need to be handled. Diagram 2 covers the possibility that z is between x and y, corresponding to case 2 in the statement of property 4.
In each case, the procedure is to write ∠z = ∠xzy as a sum or difference of two angles, α and β, and then to derive the central angle γ=∠xcy from α and β. The next step is to notice that if a triangle has two sides that are both radii of the circle, then it must be an isosceles triangle. Therefore, it must have two equal angles corresponding to the two equal sides, and the final angle can then be computed using the fact that the angles of a triangle always add up to 180º. The final step in each case is to compute angle γ in terms of α and β, and to check that the result confirms property 4.
This completes the proof.
