Trigonometric ratios, sine and cosine rules and areaEdexcel A-Level Maths: Revision notes

On the unit circle (radius $1$, centre the origin), the point at angle $\theta$, measured anticlockwise from the positive $x$-axis, has coordinates $(\cos\theta,\sin\theta)$. So $\cos\theta$ is the $x$-coordinate and $\sin\theta$ is the $y$-coordinate. Tangent is the gradient of the radius:

$\tan\theta=\frac{\sin\theta}{\cos\theta}.$

This defines the ratios for every angle, including those larger than $90^\circ$ or negative. $\tan\theta$ is undefined where $\cos\theta=0$ (at $90^\circ$ and $270^\circ$). Because $x^2+y^2=1$ on the circle, $\sin^2\theta+\cos^2\theta=1$.

The sign depends on the quadrant: all are positive for $0<\theta<90^\circ$; only $\sin$ is positive for $90^\circ<\theta<180^\circ$; only $\tan$ is positive for $180^\circ<\theta<270^\circ$; only $\cos$ is positive for $270^\circ<\theta<360^\circ$. Use the reference angle (the acute angle to the $x$-axis) for the size:

$\sin(180^\circ-\theta)=\sin\theta$, $\cos(180^\circ-\theta)=-\cos\theta$, $\sin(180^\circ+\theta)=-\sin\theta$, $\cos(360^\circ-\theta)=\cos\theta$.

Exact values: $\sin30^\circ=\frac12$, $\cos30^\circ=\frac{\sqrt3}{2}$, $\tan30^\circ=\frac1{\sqrt3}$; $\sin45^\circ=\cos45^\circ=\frac{\sqrt2}{2}$, $\tan45^\circ=1$; $\sin60^\circ=\frac{\sqrt3}{2}$, $\cos60^\circ=\frac12$, $\tan60^\circ=\sqrt3$. So $\cos150^\circ=-\frac{\sqrt3}{2}$ and $\tan150^\circ=-\frac1{\sqrt3}$.

For any triangle with sides $a,b,c$ opposite angles $A,B,C$:

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

Use it when you know a side and its opposite angle plus one other side or angle. To find an angle, use $\frac{\sin A}{a}=\frac{\sin B}{b}$. Always find the third angle first if two angles are known ($A+B+C=180^\circ$).

When you find an angle with the sine rule, $\sin\theta=\sin(180^\circ-\theta)$ gives two possible angles, $\theta$ and $180^\circ-\theta$. The second is valid only if the angles still sum to less than $180^\circ$. This happens when you are given two sides and a non-included angle, and the side opposite the given angle is shorter than the other given side.

Example: $AB=10$, $BC=7$, $A=40^\circ$: $\sin C=\frac{10\sin40^\circ}{7}=0.918$, so $C=66.7^\circ$ or $113.3^\circ$. Since $40^\circ+113.3^\circ<180^\circ$, both give a triangle, with $B=73.3^\circ$ or $26.7^\circ$ and areas $33.5$ cm$^2$ and $15.7$ cm$^2$. If the given angle were opposite the longer side, the obtuse solution would fail.

Use the cosine rule for two sides and the included angle, or for all three sides:

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

The side on the left must be opposite the angle. If $\cos A<0$ the angle is obtuse. Example: $AB=6$, $BC=10$, $B=60^\circ$ gives $AC^2=36+100-60=76$.

$\text{Area}=\frac12ab\sin C,$

where $C$ is the angle between sides $a$ and $b$. For $AB=6$, $BC=10$, $B=60^\circ$: area $=\frac12(6)(10)\sin60^\circ=15\sqrt3$. For an obtuse angle the formula still works because $\sin\theta=\sin(180^\circ-\theta)$.

Choosing a method: right angle, use SOH CAH TOA; a side and its opposite angle, use the sine rule; two sides and the included angle, or three sides, use the cosine rule; two sides and the included angle for area, use $\frac12ab\sin C$.