Trigonometric functions of any angle

The right-triangle definition limits $\sin\theta$ to $0 < \theta < \pi/2$ . To define trigonometric functions for every real angle, we use the unit circle.

Definitions

Let $\theta$ be any real number (interpreted as an angle in radians). Starting from the point $(1, 0)$ on the unit circle $x^2 + y^2 = 1$ , rotate counterclockwise through an angle $\theta$ to reach a point $P = (x, y)$ . Define:

$\sin\theta = y,\qquad \cos\theta = x,\qquad \tan\theta = \frac{y}{x} = \frac{\sin\theta}{\cos\theta} \text{ (when } \cos\theta \ne 0).$

The reciprocal functions: $\csc\theta = \frac{1}{\sin\theta},\quad \sec\theta = \frac{1}{\cos\theta},\quad \cot\theta = \frac{1}{\tan\theta} = \frac{\cos\theta}{\sin\theta}.$

This unit-circle definition agrees with the right-triangle definition for $0 < \theta < \pi/2$ but extends to all real $\theta$ .

Signs in the four quadrants

The plane is divided into four quadrants by the axes. Since $\cos\theta = x$ and $\sin\theta = y$ :

Quadrant	$x$	$y$	$\sin$	$\cos$	$\tan$
I ( $0 < \theta < \pi/2$ )	$+$	$+$	$+$	$+$	$+$
II ( $\pi/2 < \theta < \pi$ )	$-$	$+$	$+$	$-$	$-$
III ( $\pi < \theta < 3\pi/2$ )	$-$	$-$	$-$	$-$	$+$
IV ( $3\pi/2 < \theta < 2\pi$ )	$+$	$-$	$-$	$+$	$-$

Mnemonic: "All Silver Tea Cups" or "ASTC" , in quadrants I, II, III, IV the functions All, Sin, Tan, Cos (and their reciprocals) respectively are positive.

Periodicity

The unit-circle definition repeats every $2\pi$ : rotating by an additional full circle returns to the same point. Hence $\sin(\theta + 2\pi) = \sin\theta, \qquad \cos(\theta + 2\pi) = \cos\theta.$

Period of $\sin, \cos, \csc, \sec$ : $2\pi$ . Period of $\tan, \cot$ : $\pi$ (because $\tan(\theta + \pi) = \tan\theta$ ).

Domain and range

Function	Domain	Range
$\sin\theta$	$\mathbb{R}$	$[-1, 1]$
$\cos\theta$	$\mathbb{R}$	$[-1, 1]$
$\tan\theta$	$\mathbb{R} \setminus \{(2k+1)\pi/2 : k \in \mathbb{Z}\}$	$\mathbb{R}$
$\cot\theta$	$\mathbb{R} \setminus \{k\pi : k \in \mathbb{Z}\}$	$\mathbb{R}$
$\sec\theta$	$\mathbb{R} \setminus \{(2k+1)\pi/2\}$	$(-\infty, -1] \cup [1, \infty)$
$\csc\theta$	$\mathbb{R} \setminus \{k\pi\}$	$(-\infty, -1] \cup [1, \infty)$

Values at standard angles

$\theta$	$0$	$\pi/6$	$\pi/4$	$\pi/3$	$\pi/2$	$\pi$	$3\pi/2$	$2\pi$
$\sin$	$0$	$\tfrac{1}{2}$	$\tfrac{\sqrt{2}}{2}$	$\tfrac{\sqrt{3}}{2}$	$1$	$0$	$-1$	$0$
$\cos$	$1$	$\tfrac{\sqrt{3}}{2}$	$\tfrac{\sqrt{2}}{2}$	$\tfrac{1}{2}$	$0$	$-1$	$0$	$1$
$\tan$	$0$	$\tfrac{1}{\sqrt{3}}$	$1$	$\sqrt{3}$	,	$0$	,	$0$

Memorise the row for $0, \pi/6, \pi/4, \pi/3, \pi/2$ , every other value follows by reflection.

Symmetries

$\sin(-\theta) = -\sin\theta,\qquad \cos(-\theta) = \cos\theta,\qquad \tan(-\theta) = -\tan\theta.$

So $\sin, \tan, \cot, \csc$ are odd, and $\cos, \sec$ are even.

For shifts by $\pi/2$ (the so-called "complementary angle" rules): $\sin\!\left(\tfrac{\pi}{2} - \theta\right) = \cos\theta, \quad \cos\!\left(\tfrac{\pi}{2} - \theta\right) = \sin\theta, \quad \tan\!\left(\tfrac{\pi}{2} - \theta\right) = \cot\theta.$

For shifts by $\pi$ : $\sin(\pi - \theta) = \sin\theta, \quad \cos(\pi - \theta) = -\cos\theta, \quad \tan(\pi - \theta) = -\tan\theta.$

Worked examples

Example 1. Find $\sin\dfrac{7\pi}{6}$ .

$\dfrac{7\pi}{6}$ is in quadrant III ( $\pi < \dfrac{7\pi}{6} < \dfrac{3\pi}{2}$ ), where $\sin$ is negative. Reference angle: $\dfrac{7\pi}{6} - \pi = \dfrac{\pi}{6}$ . So $\sin\dfrac{7\pi}{6} = -\sin\dfrac{\pi}{6} = -\dfrac{1}{2}$ .

Example 2. Find $\cos(-300^\circ)$ .

$-300^\circ$ is coterminal with $60^\circ$ (add $360^\circ$ ). So $\cos(-300^\circ) = \cos 60^\circ = \dfrac{1}{2}$ .

Example 3. If $\sin\theta = \dfrac{3}{5}$ and $\theta$ is in quadrant II, find $\cos\theta$ and $\tan\theta$ .

Pythagorean: $\cos^2\theta = 1 - \dfrac{9}{25} = \dfrac{16}{25}$ , so $\cos\theta = \pm \dfrac{4}{5}$ . In quadrant II, $\cos < 0$ , so $\cos\theta = -\dfrac{4}{5}$ . Then $\tan\theta = \dfrac{\sin\theta}{\cos\theta} = -\dfrac{3}{4}$ .

Example 4. Compute $\tan\dfrac{11\pi}{4}$ .

$\dfrac{11\pi}{4} = 2\pi + \dfrac{3\pi}{4}$ . So $\tan\dfrac{11\pi}{4} = \tan\dfrac{3\pi}{4} = -1$ .

Example 5 (harder). Prove $\sin\!\left(\dfrac{3\pi}{2} - \theta\right) = -\cos\theta$ using the unit circle.

Rotating $\theta$ counterclockwise gives $(\cos\theta, \sin\theta)$ . Rotating by $\dfrac{3\pi}{2}$ first and then by $-\theta$ gives a point at angle $\dfrac{3\pi}{2} - \theta$ , whose $y$ -coordinate is $\sin\!\left(\dfrac{3\pi}{2} - \theta\right)$ .

Algebraically: $\sin(A - B) = \sin A \cos B - \cos A \sin B$ (next subtopic), so $\sin\!\left(\tfrac{3\pi}{2} - \theta\right) = \sin\tfrac{3\pi}{2} \cos\theta - \cos\tfrac{3\pi}{2} \sin\theta = (-1)\cos\theta - 0 \cdot \sin\theta = -\cos\theta. \qed$

Try it yourself

Compute $\cos\dfrac{5\pi}{3}$ , $\sin\dfrac{4\pi}{3}$ , $\tan\dfrac{5\pi}{4}$ .
If $\cos\theta = -\dfrac{12}{13}$ and $\theta$ in quadrant III, find $\sin\theta$ and $\tan\theta$ .
Compute $\sin(-\pi/3)$ , $\cos(-\pi/6)$ , $\tan(7\pi)$ .
Find all values of $\theta \in [0, 2\pi]$ with $\sin\theta = \dfrac{1}{2}$ .
Evaluate $\sin 750^\circ + \cos 1020^\circ$ .
Prove: $\cos(\pi - \theta) + \cos\theta = 0$ .
If $\sec\theta = 2$ and $\theta$ in quadrant IV, find $\tan\theta$ and $\sin\theta$ .
Show $\sin 7\theta + \sin 5\theta = 2\sin 6\theta \cos\theta$ in a single step (using sum-to-product).
Find the period of $f(x) = \sin\dfrac{x}{2}$ .
Prove $\tan\theta + \cot\theta = \sec\theta \csc\theta$ .
Sketch $f(x) = |\sin x|$ on $[-2\pi, 2\pi]$ .
Show that $\cos\theta \cdot \sec\theta = 1$ wherever both are defined.

Pitfalls / Tricks

The signs depend on the quadrant. Always sketch which quadrant the angle is in before assigning a sign.
$\tan\theta$ and $\sec\theta$ blow up at odd multiples of $\pi/2$ . The graphs have vertical asymptotes there.
$\sin$ and $\cos$ never exceed $1$ in magnitude. Equations like $\sin\theta = 2$ have no real solutions.
Insight. Every trigonometric question can be answered by sketching the unit circle: the $x$ -coordinate is $\cos$ , the $y$ -coordinate is $\sin$ . Master this picture and you never need to memorise sign tables again.