Skip to main content

CLP-1 Differential Calculus

Joel Feldman, Andrew Rechnitzer, Elyse Yeager

Section B.4 Cosine and Sine Laws

Subsection B.4.1 Cosine Law or Law of Cosines

The cosine law says that, if a triangle has sides of length \(a\text{,}\) \(b\) and \(c\) and the angle opposite the side of length \(c\) is \(\gamma\text{,}\) then
\begin{align*} c^2 &= a^2+b^2 - 2ab\cos\gamma \end{align*}
Observe that, when \(\gamma=\tfrac{\pi}{2}\text{,}\) this reduces to, (surpise!) Pythagoras' theorem \(c^2=a^2+b^2\text{.}\) Let's derive the cosine law.
Consider the triangle on the left. Now draw a perpendicular line from the side of length \(c\) to the opposite corner as shown. This demonstrates that
\begin{align*} c &= a \cos \beta + b \cos \alpha\\ \end{align*}

Multiply this by \(c\) to get an expression for \(c^2\text{:}\)

\begin{align*} c^2 &= ac \cos \beta + bc \cos \alpha\\ \end{align*}

Doing similarly for the other corners gives

\begin{align*} a^2 &= ac \cos \beta + ab \cos \gamma\\ b^2 &= bc \cos \alpha + ab \cos \gamma \end{align*}
Now combining these:
\begin{align*} a^2+b^2-c^2 &= (bc-bc) \cos \alpha + (ac-ac)\cos\beta + 2ab \cos \gamma\\ &= 2ab\cos \gamma \end{align*}
as required.

Subsection B.4.2 Sine Law or Law of Sines

The sine law says that, if a triangle has sides of length \(a, b\) and \(c\) and the angles opposite those sides are \(\alpha\text{,}\) \(\beta\) and \(\gamma\text{,}\) then
\begin{align*} \frac{a}{\sin \alpha} &= \frac{b}{\sin \beta} = \frac{c}{\sin \gamma}. \end{align*}
This rule is best understood by computing the area of the triangle using the formula \(A = \frac{1}{2}ab\sin\theta\) of Appendix A.10. Doing this three ways gives
\begin{align*} 2A &= bc \sin \alpha\\ 2A &= ac \sin \beta\\ 2A &= ab \sin \gamma \end{align*}
Dividing these expressions by \(abc\) gives
\begin{align*} \frac{2A}{abc} &= \frac{\sin \alpha}{a} = \frac{\sin\beta}{b} = \frac{\sin \gamma}{c} \end{align*}
as required.
<Prev^TopNext>
Feedback