Radians, Unit Circle, Trig Identities & 3D Applications
Radians provide a natural unit for angle measurement. One radian is the angle subtended at the centre of a circle by an arc equal in length to the radius.
For a circle of radius $r$ and central angle $\theta$ (in radians):
$$\text{Arc length: } s = r\theta \qquad \text{Sector area: } A = \frac{1}{2}r^2\theta$$
Conversion: $180° = \pi \text{ rad}$, so to convert degrees to radians, multiply by $\dfrac{\pi}{180}$.
On the unit circle (radius 1), a point at angle $\theta$ from the positive $x$-axis has coordinates $(\cos\theta, \sin\theta)$.
| $\theta$ (rad) | $\theta$ (deg) | $\sin\theta$ | $\cos\theta$ | $\tan\theta$ |
|---|---|---|---|---|
| $0$ | $0°$ | $0$ | $1$ | $0$ |
| $\pi/6$ | $30°$ | $\frac{1}{2}$ | $\frac{\sqrt{3}}{2}$ | $\frac{1}{\sqrt{3}}$ |
| $\pi/4$ | $45°$ | $\frac{\sqrt{2}}{2}$ | $\frac{\sqrt{2}}{2}$ | $1$ |
| $\pi/3$ | $60°$ | $\frac{\sqrt{3}}{2}$ | $\frac{1}{2}$ | $\sqrt{3}$ |
| $\pi/2$ | $90°$ | $1$ | $0$ | undefined |
The CAST diagram shows in which quadrants trig ratios are positive:
Pythagorean identity: $\sin^2\theta + \cos^2\theta = 1$
Tangent: $\tan\theta = \dfrac{\sin\theta}{\cos\theta}$
Double angle formulas:
$\sin 2\theta = 2\sin\theta\cos\theta$
$\cos 2\theta = \cos^2\theta - \sin^2\theta = 2\cos^2\theta - 1 = 1 - 2\sin^2\theta$
$\tan 2\theta = \dfrac{2\tan\theta}{1-\tan^2\theta}$
For a triangle with sides $a, b, c$ opposite to angles $A, B, C$ respectively:
$$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$$
Use when you know: (i) two angles and one side, or (ii) two sides and a non-included angle (beware the ambiguous case!).
$$c^2 = a^2 + b^2 - 2ab\cos C$$
$$\cos C = \frac{a^2 + b^2 - c^2}{2ab}$$
Use when you know: (i) two sides and the included angle, or (ii) all three sides.
$$\text{Area} = \frac{1}{2}ab\sin C$$
When you know two sides and a non-included angle, there may be 0, 1, or 2 possible triangles. Always check whether a second solution exists by considering the supplementary angle $\pi - A$.
Three-dimensional problems require identifying the correct plane, right triangle, or triangle within the 3D figure.
Distance between $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$:
$$d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}$$
A sector has radius $8$ cm and central angle $\dfrac{2\pi}{5}$ radians. Find the arc length and area of the sector.
In triangle $ABC$, $a = 7$, $b = 10$, and $A = 35°$. Find the two possible values of angle $B$.
A vertical pole $PQ$ of height 6 m stands at corner $Q$ of a rectangular field $QRST$, where $QR = 8$ m and $QS = 10$ m. Find the angle that the line $PR$ makes with the base plane $QRST$.
Q1. A sector of a circle has area $24\pi$ cm² and radius $6$ cm. Find the central angle in radians and the arc length.
$A = \frac{1}{2}r^2\theta \implies 24\pi = \frac{1}{2}(36)\theta \implies \theta = \frac{24\pi}{18} = \frac{4\pi}{3}$ rad.
Arc length: $s = r\theta = 6 \times \frac{4\pi}{3} = 8\pi \approx 25.1$ cm.
Q2. Prove that $\dfrac{\sin 2\theta}{1 + \cos 2\theta} = \tan\theta$.
LHS $= \dfrac{2\sin\theta\cos\theta}{1 + (2\cos^2\theta - 1)} = \dfrac{2\sin\theta\cos\theta}{2\cos^2\theta} = \dfrac{\sin\theta}{\cos\theta} = \tan\theta$ = RHS ✓
Q3. In triangle $PQR$, $PQ = 11$ cm, $PR = 8$ cm, $QR = 9$ cm. Find angle $P$ to the nearest degree.
Using cosine rule: $QR^2 = PQ^2 + PR^2 - 2(PQ)(PR)\cos P$
$81 = 121 + 64 - 2(11)(8)\cos P = 185 - 176\cos P$
$\cos P = \dfrac{185 - 81}{176} = \dfrac{104}{176} \approx 0.5909$
$P = \arccos(0.5909) \approx 54°$
Q4. Find the exact value of $\sin\left(\dfrac{7\pi}{6}\right)$ using the unit circle.
$\frac{7\pi}{6} = \pi + \frac{\pi}{6}$, so the angle is in Quadrant III (between $\pi$ and $\frac{3\pi}{2}$).
In Quadrant III, $\sin$ is negative.
Reference angle: $\frac{7\pi}{6} - \pi = \frac{\pi}{6}$, and $\sin\frac{\pi}{6} = \frac{1}{2}$.
Therefore $\sin\frac{7\pi}{6} = -\dfrac{1}{2}$.