Trigonometry

Introduction

This chapter studies the relationships between the sides and angles of a triangle, and uses them to "solve" a triangle, that is, to find its unknown sides and angles from the ones that are given. Fix the standard notation once and for all: in triangle $ABC$ the angles are $A, B, C$ and the sides opposite to them are $a, b, c$, so $a = BC$, $b = CA$, $c = AB$. The three angles satisfy

$$ \begin{aligned} A + B + C &= \pi \ &= 180^{\circ} \end{aligned} $$

a fact used constantly. Throughout, $s = \dfrac{a+b+c}{2}$ is the semi-perimeter and $\Delta$ denotes the area.

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Sine law

In any triangle the sides are proportional to the sines of the opposite angles:

$$ \begin{aligned} \frac{a}{\sin A} &= \frac{b}{\sin B} \ &= \frac{c}{\sin C} \ &= 2R \end{aligned} $$

where $R$ is the radius of the circumscribed circle (circumradius). Equivalently $a : b : c = \sin A : \sin B : \sin C$. The sine law is the natural tool when you are given two angles and one side (ASA or AAS), or two sides and an angle opposite one of them (SSA).

Worked example. In triangle $ABC$, $A = 45^{\circ}$, $B = 60^{\circ}$, $a = 10$. Find $b$. Since $\dfrac{a}{\sin A} = \dfrac{b}{\sin B}$,

$$ \begin{aligned} b &= \frac{a\sin B}{\sin A} = \frac{10\sin 60^{\circ}}{\sin 45^{\circ}} \ &= \frac{10\cdot\frac{\sqrt3}{2}}{\frac{\sqrt2}{2}} = \frac{10\sqrt3}{\sqrt2} = 5\sqrt6. \end{aligned} $$

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Cosine law

The cosine law relates one angle to all three sides, and is the tool of choice when you know all three sides (SSS), or two sides and the included angle (SAS):

$$ \begin{aligned} a^2 &= b^2 + c^2 - 2bc\cos A, \ b^2 &= c^2 + a^2 - 2ca\cos B, \ c^2 &= a^2 + b^2 - 2ab\cos C. \end{aligned} $$

Rearranged to give an angle,

$$\cos A = \frac{b^2 + c^2 - a^2}{2bc}.$$

When $A = 90^{\circ}$, $\cos A = 0$ and the law reduces to Pythagoras' theorem $a^2 = b^2 + c^2$, so the cosine law is a generalisation of it.

Worked example. A triangle has $b = 5$, $c = 8$ and included angle $A = 60^{\circ}$. Find $a$:

$$ \begin{aligned} a^2 &= 5^2 + 8^2 - 2\cdot 5\cdot 8\cos 60^{\circ} \ &= 25 + 64 - 80\cdot\tfrac12 \ &= 89 - 40 = 49, \end{aligned} $$

so $a = 7$.

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