Radian measure, arc length and exact valuesEdexcel A-Level Maths: Revision notes

One radian is the angle at the centre of a circle subtended by an arc equal in length to the radius. A full turn is $2\pi$ radians, so $\pi$ radians $=180^\circ$. To convert, multiply degrees by $\frac{\pi}{180}$ and radians by $\frac{180}{\pi}$. $30^\circ=\frac{\pi}{6},\quad45^\circ=\frac{\pi}{4},\quad60^\circ=\frac{\pi}{3},\quad90^\circ=\frac{\pi}{2},\quad150^\circ=\frac{5\pi}{6}.$ Leave angles as multiples of $\pi$ when exact answers are needed. The formulae below only work in radians, so check the angle mode on your calculator.

For a sector of radius $r$ and angle $\theta$ radians: $s=r\theta,\qquad A=\tfrac12r^2\theta.$ The perimeter of a sector is $2r+r\theta$, which includes the two straight radii. Example: $r=6$, $\theta=\frac{2\pi}{3}$ gives $s=4\pi$, $A=12\pi$ and perimeter $12+4\pi$. If the perimeter and radius are given, form an equation in $\theta$: with $r=5$ and perimeter $16$, $10+5\theta=16$ so $\theta=1.2$.

A segment is the region between a chord and an arc. Its area is the sector minus the triangle: $A_{\text{segment}}=\tfrac12r^2\theta-\tfrac12r^2\sin\theta=\tfrac12r^2(\theta-\sin\theta).$ Example: $r=5$, $\theta=1.2$: sector $15$, triangle $\frac12(25)\sin1.2=11.65$, segment $3.35$ cm$^2$. For a composite shape, add or subtract sectors and triangles, and use the cosine rule when a chord is needed: $AB^2=2r^2-2r^2\cos\theta$.

You must know these exact values for $\theta=0,\frac{\pi}{6},\frac{\pi}{4},\frac{\pi}{3},\frac{\pi}{2}$: $\sin\theta:\ 0,\ \frac12,\ \frac{\sqrt2}{2},\ \frac{\sqrt3}{2},\ 1$ $\cos\theta:\ 1,\ \frac{\sqrt3}{2},\ \frac{\sqrt2}{2},\ \frac12,\ 0$ The sine values are $\frac{\sqrt0}{2},\frac{\sqrt1}{2},\frac{\sqrt2}{2},\frac{\sqrt3}{2},\frac{\sqrt4}{2}$, and cosine is the same list reversed. Derive the $\frac{\pi}{6}$ and $\frac{\pi}{3}$ values from an equilateral triangle of side $2$ cut in half, and $\frac{\pi}{4}$ from a right-angled isosceles triangle.

Use $\tan\theta=\frac{\sin\theta}{\cos\theta}$: $\tan0=0,\quad\tan\frac{\pi}{6}=\frac{1}{\sqrt3}=\frac{\sqrt3}{3},\quad\tan\frac{\pi}{4}=1,\quad\tan\frac{\pi}{3}=\sqrt3.$ $\tan\frac{\pi}{2}$ is undefined because $\cos\frac{\pi}{2}=0$. Rationalise surds when asked for the simplest form: $\frac{1}{\sqrt3}=\frac{\sqrt3}{3}$.

Use the unit circle and symmetry to find values at angles such as $\frac{2\pi}{3},\frac{3\pi}{4},\frac{5\pi}{6},\pi,\frac{7\pi}{6},\frac{3\pi}{2}$. Find the acute related angle, then decide the sign from the quadrant (sine positive in the first two, cosine in the first and fourth, tangent in the first and third). $\sin\frac{5\pi}{6}=\frac12,\quad\cos\frac{5\pi}{6}=-\frac{\sqrt3}{2},\quad\tan\frac{5\pi}{6}=-\frac{\sqrt3}{3}.$ Also $\sin(\pi-\theta)=\sin\theta$, $\cos(\pi-\theta)=-\cos\theta$ and $\sin(\pi+\theta)=-\sin\theta$.