Trigonometry and Triangles

Law of Sines

Consider this triangle.

Draw a line vertically down from \(B\). Label the length of this line as \(h\) (for height).

Now \(\sin(\pi-\alpha)=h/c=\sin\alpha\)

But it’s also the case that \(\sin\gamma=h/a\).

Equate \(h\) in the two equations and you’ll get the Law of Sines:

\begin{equation*} \frac{a}{\sin \alpha}=\frac{b}{\sin \beta}=\frac{c}{\sin \gamma} \end{equation*}

Law of Cosines

Consider the same triangle as above. The coordinates of \(B\) are \((c\cos\alpha, c\sin\alpha)\). The coordinates of \(C\) are \((b,0)\). Use the distance formula to write the expression for the length of \(BC\):

\begin{equation*} \ |\overline{B C}|^{2}=(c \cos \alpha-b)^{2}+(c \sin \alpha-0)^{2} \end{equation*}

From here you’ll get:

\begin{equation*} a^{2}=b^{2}+c^{2}-2 b c \cos \alpha \end{equation*}

Similarly:

\begin{equation*} \begin{array}{l}b^{2}=c^{2}+a^{2}-2 c a \cos \beta \\ c^{2}=a^{2}+b^{2}-2 a b \cos \gamma\end{array} \end{equation*}

This derivations is very dissatisfying. It’s as if you just place the triangle on the most convenient coordinates to get the result. No intuition.

Let’s try again.

The cosine law looks a lot like Pythagora’s Theorem:

\begin{equation*} a^{2}=b^{2}+c^{2} \end{equation*}

Is this law a generalization of Pythagora’s Theorem?

Let’s take the same triangle as above but set \(\alpha=\pi/2\). Now let’s change \(\alpha\) a little. [1] More importantly, let’s fix the lengths \(b\) and \(c\). The question now is: What is the new \(a\)?

The new coordinates for \(B\) are \((c\cos\alpha,c\sin\alpha)\). The coordinates of \(C\) are \((b,0)\). Now use the distance formula and you’ll get the result.

Admittedly, this is the same derivation, but somehow it’s easier for me to understand if I view it as starting from a right triangle and fixing \(b\) and \(c\).

The Law of Tangents

From the Law of Sines, we have:

\begin{equation*} \frac{a}{b}=\frac{\sin \alpha}{\sin \beta} \end{equation*}

Applying componendo-dividendo, we have:

\begin{equation*} \frac{a-b}{a+b}=\frac{\sin \alpha-\sin \beta}{\sin \alpha+\sin \beta}= \frac{2 \cos \frac{\alpha+\beta}{2} \sin \frac{\alpha-\beta}{2}}{2 \sin \frac{\alpha+\beta}{2} \cos \frac{\alpha-\beta}{2}} \end{equation*}

\begin{equation*} \frac{a-b}{a+b}=\frac{\tan \frac{\alpha-\beta}{2}}{\tan \frac{\alpha+\beta}{2}} \end{equation*}

Half Angle Formulae

In any triangle:

\begin{equation*} \sin \frac{\alpha}{2}=\sqrt{\frac{(s-b)(s-c)}{b c}} \end{equation*}

Proof:

\begin{equation*} \begin{aligned} & 2 \sin ^{2} \frac{\alpha}{2}=1-\cos \alpha \\ \therefore \quad & 2 \sin ^{2} \frac{\alpha}{2}=1- \frac{b^{2}+c^{2}-a^{2}}{2 b c} \\ \therefore \quad & 2 \sin ^{2} \frac{\alpha}{2}=\frac{a^{2}-\left(b^{2}+c^{2}-2 b c\right)}{2 b c}=\frac{a^{2}-(b-c)^{2}}{2 b c} \\ \therefore \quad & \sin ^{2} \frac{\alpha}{2}=\frac{(a+b-c)(a-b+c)}{4 b c} \\ \therefore \quad & \sin ^{2} \frac{\alpha}{2}=\frac{2(s-c) .2(s-b)}{4 b c} \end{aligned} \end{equation*}

You can similarly show:

\begin{equation*} \cos \frac{\alpha}{2}=\sqrt{\frac{s(s-a)}{b c}} \end{equation*}

Also:

\begin{equation*} \tan \frac{\alpha}{2}=\sqrt{\frac{(s-b)(s-c)}{s(s-a)}} \end{equation*}

You can obtain this by dividing the two previous identities.

Footnotes

[1]	It really doesn’t need to be “a little”.