3.5 The unit circle and exact valuesIB Maths: Analysis and Approaches SL: Revision notes

Draw a circle of radius 1 centred at the origin (the unit circle). Rotate a radius anticlockwise from the positive $x$-axis through an angle $\theta$. The point where it meets the circle is $(\cos\theta,\ \sin\theta).$ So $\cos\theta$ is the $x$-coordinate and $\sin\theta$ is the $y$-coordinate. This defines sine and cosine for any angle, including obtuse, reflex and negative angles (negative angles are measured clockwise). Since the point is on $x^2+y^2=1$, both values lie between $-1$ and $1$.

The signs follow the coordinates: first quadrant both positive; second quadrant $\cos<0$, $\sin>0$; third quadrant both negative; fourth quadrant $\cos>0$, $\sin<0$. Symmetry of the circle gives the key relationships:

• Reflection in the $x$-axis: $\cos(-x)=\cos x$, $\sin(-x)=-\sin x$.
• Reflection in the $y$-axis: $\sin(\pi-x)=\sin x$, $\cos(\pi-x)=-\cos x$.
• Half-turn about the origin: $\sin(\pi+x)=-\sin x$, $\cos(\pi+x)=-\cos x$.
• $2\pi-x$ is the same point as $-x$. So every angle can be linked to an acute reference angle with the sign fixed by the quadrant.

$\tan\theta=\frac{\sin\theta}{\cos\theta},\quad\cos\theta\ne0.$ Tangent is the gradient of the radius $OP$, so the line through the origin at angle $\theta$ to the positive $x$-axis is $y=x\tan\theta.$ $\tan\theta$ is positive in the first and third quadrants and negative in the second and fourth. It is undefined at $\theta=\frac{\pi}{2}$, $\frac{3\pi}{2}$, where the line is vertical. Example: $\tan\frac{2\pi}{3}=\frac{\sqrt3/2}{-1/2}=-\sqrt3$, so $OP$ is $y=-\sqrt3x$.

Learn these (the booklet does not give them):

• $\theta=0$: $\sin=0$, $\cos=1$, $\tan=0$.
• $\theta=\frac{\pi}{6}$: $\sin=\frac12$, $\cos=\frac{\sqrt3}{2}$, $\tan=\frac{1}{\sqrt3}$.
• $\theta=\frac{\pi}{4}$: $\sin=\cos=\frac{\sqrt2}{2}$, $\tan=1$.
• $\theta=\frac{\pi}{3}$: $\sin=\frac{\sqrt3}{2}$, $\cos=\frac12$, $\tan=\sqrt3$.
• $\theta=\frac{\pi}{2}$: $\sin=1$, $\cos=0$, $\tan$ undefined.

For a multiple such as $\frac{7\pi}{6}$: find the reference angle ($\frac{\pi}{6}$), the quadrant (third), then the sign: $\sin\frac{7\pi}{6}=-\frac12$.

When you know two sides and a non-included angle, using the sine rule to find an angle can give two answers, because $\sin C=\sin(180^\circ-C)$ (the unit circle symmetry $\sin(\pi-x)=\sin x$). Example: $AB=10$, $AC=6$, $B=30^\circ$. Then $\sin C=\frac56$, so $C=56.4^\circ$ or $123.6^\circ$. Check each: $30^\circ+123.6^\circ<180^\circ$, so both triangles exist. The cosine rule shows the same thing algebraically: with $BC=x$, $36=100+x^2-10\sqrt3x$ is a quadratic with two positive roots, $5\sqrt3\pm\sqrt{11}$. The second triangle exists only if the obtuse angle plus the given angle is less than $180^\circ$.