Trigonometry

Trig functions

The unit circle is a circle with radius 1 centred at the origin (possibly NIS but a useful model regardless).

For an angle $\theta$ drawn anticlockwise from the $x$ axis:

• $\cos \theta =\frac{OB}{OA}$ and $OA=1$, so $x=OB=\cos \theta$.
• $\sin \theta =\frac{AB}{OA}$ and $OA=1$, so $y=AB=\sin \theta$.

There are other trig functions:

$$\begin{array}{rl}\tan \theta =\frac{\sin \theta }{\cos \theta }& \cot \theta =\frac{1}{\tan \theta }=\frac{\cos \theta }{\sin \theta }\\ & \\ \sec \theta =\frac{1}{\cos \theta }& \mathrm{cosec}\theta =\frac{1}{\sin \theta }\end{array}$$

Graphs

Graph of $\cos \theta$

The $\cos \theta$ graph is even (symmetric about $x=0$) and continues indefinitely with a period of $2\pi$.

Graph of $\sin \theta$

The $\sin \theta$ graph is odd (rotational symmetry order 2 about the origin) and also continues indefinitely with a period of $2\pi$.

Graph of $\tan \theta$

$\tan \theta =\frac{\sin \theta }{\cos \theta }$, so the graph passes through the origin. Both $\sin \theta$ and $\cos \theta$ are positive in the first quadrant, so $\tan \theta$ is positive for $0<x<\frac{\pi }{2}$. As $\cos \theta$ gets closer to 0 as $x$ approaches $\frac{\pi }{2}$, $\tan \theta$ gets larger and larger, with an asymptote at $\theta =\frac{\pi }{2}$. The graph continues indefinitely with a period of $\pi$.

Trig identities

The two most basic trigonometric identities are:

$$\begin{array}{r}{\sin }^2\theta +{\cos }^2\theta =1\\ \\ \tan \theta =\frac{\sin \theta }{\cos \theta }\end{array}$$

Dividing the first identity through by ${\sin }^2\theta$ or ${\cos }^2\theta$ gives:

$$\begin{array}{rl}{\cot }^2\theta +1& ={\mathrm{cosec}}^2\theta \\ & \\ {\tan }^2\theta +1& ={\sec }^2\theta \\ & \end{array}$$

Sine rule

The sine rule

$$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$$

When using the sine rule to find angles, sometimes there may be two possible answers, $A$ and $180°-A$.

Cosine rule

$$a^2=b^2+c^2-2bc\cos A$$

Area of a triangle

The area of a triangle with sides $a$ and $b$ and an angle between them $C$ is given by:

$$\frac{1}{2}ab\sin C$$

Radians

Angles can be measured in degrees or radians, with $2\pi (\approx 6.28)$ radians in a circle ($360°$). Thus, to convert between them:

• To convert from radians to degrees, multiply by $\frac{180°}{\pi }$
• To convert from degrees to radians, multiply by $\frac{pi}{180°}$.

Radians and geometry

The length of an arc of a circle with radius $r$ over angle $\theta$ subtended at the centre, in radians, is given by:

$$l=r\theta$$

The area of a sector of circle with radius $r$ over angle $\theta$ subtended at the centre, in radians, is given by:

$$A=\frac{1}{2}{r}^2\theta$$

Small angle approximations

The small angle approximations can be used for small angles measured in radians:

$$\begin{array}{rl}\sin \theta & \approx \theta \\ \cos \theta & \approx 1-\frac{{\theta }^2}{2}\\ \tan \theta & \approx \theta \end{array}$$

Compound angle formulae

Geometric derivation

The main derivation for compound angle formulae is for $\cos (\alpha +\beta )$, with the other formulae following from this one. IS: geometric proof of the compound-angle formulae.

Consider $△ABC$ as below, with $AH$ as an altitude to side $BC$.

The area of $△ACH$ can be calculated:

$$\begin{array}{rl}△ACH& =\frac{1}{2}(AC)(AH)\sin \alpha \\ & =\frac{1}{2}(AC)(AB\cos \beta )\sin \alpha \\ & =\frac{1}{2}(AC)(AB)\sin \alpha \cos \beta \end{array}$$

The area of $△ABH$ can also be calculated:

$$\begin{array}{rl}△ABH& =\frac{1}{2}(AB)(AH)\sin \beta \\ & =\frac{1}{2}(AB)(AC\cos \alpha )\sin \beta \\ & =\frac{1}{2}(AB)(AC)\cos \alpha \sin \beta \end{array}$$

By considering the area of $△ABC$ as a whole:

$$\begin{array}{rl}△ABC& =△ACH+△ABH\\ \frac{1}{2}(AB)(AC)\sin (\alpha +\beta )& =\frac{1}{2}(AC)(AB)\sin \alpha \cos \beta +\frac{1}{2}(AB)(AC)\sin \beta \cos \alpha \\ \sin (\alpha +\beta )& =\sin \alpha \cos \beta +\sin \beta \cos \alpha \end{array}$$

Giving the final result.

This result can be manipulated to derive the 3 other compound-angle identities for $\sin$ and $\cos$, which we leave a proof for below. These identities can be further manipulated to derive 2 additional compound-angle identities for $\tan$. The below 6 identities are given:

$$\begin{array}{rl}\sin (\alpha \pm \beta )& =\sin \alpha \cos \beta \pm \sin \beta \cos \alpha \\ & \\ \cos (\alpha \pm \beta )& =\cos \alpha \cos \beta \mp \sin \alpha \sin \beta \\ & \\ \tan (\alpha \pm \beta )& =\frac{\tan \alpha \pm \tan \beta }{1\mp \tan \alpha \tan \beta }\end{array}$$

Phase shift form

An expression of the form $a\sin \theta +b\cos \theta$ can also be written in the forms:

$$R\sin (\theta \pm \alpha )\text{or}R\cos (\theta \pm \alpha )$$

for suitably chosen $R$ and $\alpha$. $R$ represents the amplitude of the resulting expression, while $\theta$ is a phase shift.

To rewrite an expression into these forms:

• Start with the target form and apply the compound-angle identity
• Compare coefficients on the $\sin \theta$ and $\cos \theta$ terms
• The amplitude is always given by $R=\sqrt{a^2+b^2}$
• By dividing $\sin \alpha$ by $\cos \alpha$, $\tan \alpha$ and thus $\alpha$ can be found, taking care with signs
• Write the final expression with substituted values for $R$ and $\alpha$.

Proofs

The cosine rule

Consider $△ABC$ such that $AC=b$, $CB=a$, and $\angle ACB=C$. By SAS congruency, all triangles with these properties must be congruent, so $BC$ can only have one length.

$AH$ is an altitude drawn to $CB$. The length of $CH$ can be calculated using basic trigonometry in right angle triangle $△AHC$. $\cos C=\frac{CH}{b}$, so $CH=b\cos C$.

With $CH$ known, $AH$ can be calculated using Pythagoras. $A{H}^2=b^2-(b\cos C{)}^2$ $HB=CB-CH=a-b\cos C$ With $HB$ and $AH$ known, $AB$ can be calculated using Pythagoras.

$$\begin{array}{rl}A{B}^2& =A{H}^2+H{B}^2\\ & =b^2-(b\cos C{)}^2+a^2+(b\cos C{)}^2-2ab\cos C\\ & =b^2+a^2-2ab\cos C\\ c^2& =a^2+b^2-2ab\cos C\end{array}$$

Or alternatively:

$$a^2=b^2+c^2-2bc\cos (A)$$

The sine rule

Here, we know $\angle A$ and $\angle B$ and the side $C$ between them. We draw in the circumcircle of the triangle, making $AE$ a diameter of the circumcircle, which has radius $R$.

Because $\angle AEB$ and $\angle ACB$ subtend the same arc, they are equal.

$$\begin{array}{rl}\sin C& =\sin \angle AEB=\frac{AB}{BE}=\frac{c}{2R}\\ 2R& =\frac{c}{\sin C}\end{array}$$

As there is nothing special about side $AB$, we can use the symmetry to write:

$$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R$$

Sine and cosine compound angle formulae

Starting with $\sin(\alpha + \beta) = \sin\alpha \cos\beta - \sin\beta\cos\alpha$$, we can apply some manipulations to find the three other variations.

To find $\sin (\alpha -\beta )$:

$$\begin{array}{rl}\sin (\alpha -\beta ))& =\sin (\alpha +(-\beta ))\\ & =\sin (\alpha )\cos (-\beta )+\sin (-\beta )\cos (\alpha )\\ & =\sin (\alpha )\cos (\beta )-\sin (\beta )\cos (\alpha )\end{array}$$

Where on the third line we apply $\cos (-\theta )=\cos (\theta )$ and $\sin (-\theta )=-\sin (\theta )$.

To find $\cos (\alpha +\beta )$:

$$\begin{array}{rl}\cos (\alpha +\beta )& =\sin \left(\left(\frac{\pi }{2}-\alpha \right)-\beta \right)\\ & =\sin \left(\frac{\pi }{2}-\alpha \right)\cos (\beta )-\sin (\beta )\cos \left(\frac{\pi }{2}-\alpha \right)\\ & =\cos (\alpha )\cos (\beta )-\sin (\beta )\sin (\alpha )\end{array}$$

Where on the third line we apply $\sin (\theta )=\cos \left(\frac{\pi }{2}-\theta \right)$ and $\cos (\theta )=\sin \left(\frac{\pi }{2}-\theta \right)$

To find $\cos (\alpha -\beta )$:

$$\begin{array}{rl}\cos (\alpha -\beta )& =\sin \left(\frac{\pi }{2}-(\alpha -\beta )\right)\\ & =\sin \left(\left(\frac{\pi }{2}-\alpha \right)+\beta \right)\\ & =\sin \left(\frac{\pi }{2}-\alpha \right)\cos (\beta )+\sin (\beta )\cos \left(\frac{\pi }{2}-\alpha \right)\\ & =\cos (\alpha )\cos (\beta )+\sin (\beta )\sin (\alpha )\end{array}$$

Giving us all four addition formulae for $\sin$ and $\cos$:

$$\begin{array}{rl}\sin (\alpha +\beta )& =\sin (\alpha )\cos (\beta )+\sin (\beta )\cos (\alpha )\\ \sin (\alpha -\beta )& =\sin (\alpha )\cos (\beta )-\sin (\beta )\cos (\alpha )\\ \cos (\alpha +\beta )& =\cos (\alpha )\cos (\beta )-\sin (\beta )\sin (\alpha )\\ \cos (\alpha -\beta )& =\cos (\alpha )\cos (\beta )+\sin (\beta )\sin (\alpha )\end{array}$$

Tangent compound angle formulae

From the $\sin$ and $\cos$ compound angle formulae, we can derive the compound angle formulae for $\tan$:

$$\begin{array}{rl}\tan (\alpha +\beta )& =\frac{\sin (\alpha +\beta )}{\cos (\alpha +\beta )}\\ & \\ & =\frac{\sin (\alpha )\cos (\beta )+\sin (\beta )\cos (\alpha )}{\cos (\alpha )\cos (\beta )-\sin (\beta )\sin (\alpha )}\\ & \\ & =\frac{\tan (\alpha )+\tan (\beta )}{1-\tan (\alpha )\tan (\beta )}\end{array}$$

Where on the third line we divide the numerator and denominator by $\cos (\alpha )\cos (\beta )$.

Similarly:

$$\begin{array}{rl}\tan (\alpha -\beta )& =\frac{\sin (\alpha -\beta )}{\cos (\alpha -\beta )}\\ & \\ & =\frac{\sin (\alpha )\cos (\beta )-\sin (\beta )\cos (\alpha )}{\cos (\alpha )\cos (\beta )+\sin (\beta )\sin (\alpha )}\\ & \\ & =\frac{\tan (\alpha )-\tan (\beta )}{1+\tan (\alpha )\tan (\beta )}\end{array}$$