The sine and cosine theorem

The sine theorem is a theorem that - unlike ordinary goniometric functions - holds in a general triangle. It gives us the relationship between side lengths and angles. The cosine theorem also holds in a general triangle, and its special case is the Pythagorean theorem.

Sine's theorem

The Sine Theorem tells us that the ratio of all side lengths and the sine values of their opposite angles is constant in a given general (!) triangle. We write it as follows:

$$\frac{a}{\sin\alpha}=\frac{b}{\sin\beta}=\frac{c}{\sin\gamma}=2r,$$

where r is the radius of the circle described and a, b and c are the lengths of the sides of the triangle. We can also rewrite the previous equality in the form:

$$\frac{a}{b}=\frac{\sin\alpha}{\sin\beta}, \frac{b}{c}=\frac{\sin\beta}{\sin\gamma}, \frac{c}{a}=\frac{\sin\gamma}{\sin\alpha}$$

Cosine Theorem

The cosine theorem also holds in a general triangle, as does the sine theorem. Cosine's theorem reads:

$$\begin{eqnarray} a^2 &=& b^2 + c^2 - 2 b c \cdot \cos \alpha\\ b^2 &=& c^2 + a^2 - 2 c a \cdot \cos \beta\\ c^2 &=& a^2 + b^2 - 2 a b \cdot \cos \gamma \end{eqnarray}$$

Note that if one of the angles is right, i.e., has a magnitude of 90 degrees, then the last part of the formula disappears and we get the form of the Pythagorean Theorem. Because if $\alpha=90^{\circ}$, then cos(α) = 0 and so we get:

$$\begin{eqnarray} a^2 &=& b^2 + c^2 - 2 b c \cdot \cos \alpha\\ a^2 &=& b^2 + c^2 - 2 b c \cdot 0\\ a^2 &=& b^2 + c^2 \end{eqnarray}$$

Motivation

Consider a triangle ABC with side c of length |c| = 9. The sizes of the angles are: $\alpha=40^{\circ}, \beta=80^{\circ}, \gamma=60^{\circ}$. The question is, what is the length of the remaining two sides? At this point, we can't make do with the usual goniometric functions, because they work in a right triangle, which this is definitely not. Figure:

At this point, we'll have to use other routes. One of them is just the sine theorem. Which way do we use it? The sine theorem tells us that:

$$\frac{|a|}{\sin\alpha}=\frac{|b|}{\sin\beta}=\frac{|c|}{\sin\gamma}$$

We know all the angles and the length of the side c. So we compute the side b from the equation

$$\frac{|b|}{\sin\beta}=\frac{|c|}{\sin\gamma}$$

Here we have to isolate the side length b. We do an equivalent adjustment of the equations and multiply the equation by the sine of the angle beta. We get:

$$|b|=\frac{|c|\cdot\sin\beta}{\sin\gamma}$$

We know or can calculate all the expressions on the first side. So we add:

$$|b|=\frac{9\cdot0{,}985}{0{,}866}=10{,}23.$$

In exactly the same way, we calculate the remaining side of a:

$$\frac{|a|}{\sin\alpha}=\frac{|c|}{\sin\gamma}$$

We isolate |a| by multiplying the equation by the sine of the angle alpha.

$$|a|=\frac{|c|\cdot\sin\alpha}{\sin\gamma}$$

Let's calculate the result:

$$|a|=\frac{9\cdot0{,}642}{0{,}866}=6{,}672$$

Deriving the sine theorem

Why does the sine equality hold? Consider an ordinary triangle ABC, where we still plot the height to the side c.

The question is, what is the length of the side CP_c. With the height, we have a triangle ABC divided into two right triangles, which we can use to express the length of the height. So let's try to express the length of the side CP_c using the two triangles. It is true that the sine of angle alpha is equal:

$$\sin(\alpha)=\frac{|CP_c|}{|AC|}$$

From here we express |CP_c|:

$$|CP_c|=\sin(\alpha)\cdot|AC|=\sin(\alpha)\cdot |b|.$$

Now we express the length of the side CP_c using the second triangle, using the angle beta:

$$\sin(\beta)=\frac{|CP_c|}{|BC|}$$

From here we again isolate |CP_c|:

$$|CP_c|=\sin(\beta)\cdot|BC|=\sin(\beta)\cdot |a|$$

So at this point we have the length of the height expressed by two formulas:

$$|CP_c|=\sin(\alpha)\cdot |b|=\sin(\beta)\cdot |a|,$$

so we get the equality

$$\sin(\alpha)\cdot |b|=\sin(\beta)\cdot |a|.$$

That's a bit closer to what the sine theorem is telling us. We divide the whole equation by sine alpha times sine beta:

$$\begin{eqnarray} \sin(\alpha)\cdot |b|&=&\sin(\beta)\cdot |a|\qquad/:\sin(\alpha)\cdot\sin(\beta)\\ \frac{\sin(\alpha)\cdot |b|}{\sin(\alpha)\cdot\sin(\beta)}&=&\frac{\sin(\beta)\cdot |a|}{\sin(\alpha)\cdot\sin(\beta)}\qquad/\mbox{ shorten the siny }\\ \frac{|b|}{\sin(\beta)}&=&\frac{|a|}{\sin(\alpha)} \end{eqnarray}$$

We proved part of the Sine Theorem in the previous procedure, however the proof for the other combinations of sides is identical.