Trig Ratios in Radians

Changing the unit of an angle doesn’t change the angle, so it doesn’t change its trig ratios either: $\sin\tfrac{\pi}{6}$ is the same number as $\sin 30^\circ$. This page takes everything you know about trig ratios of any angle and reciprocal ratios and moves it into radians, which is the unit for the rest of this course.

Key ideas

Same ratios, new unit

For an angle $\theta$ in standard position with a point $P(x, y)$ on its terminal arm, and $r = \sqrt{x^2 + y^2}$:

$$\sin\theta = \frac{y}{r} \qquad \cos\theta = \frac{x}{r} \qquad \tan\theta = \frac{y}{x}$$ $$\csc\theta = \frac{1}{\sin\theta} = \frac{r}{y} \qquad \sec\theta = \frac{1}{\cos\theta} = \frac{r}{x} \qquad \cot\theta = \frac{1}{\tan\theta} = \frac{x}{y}$$

On the unit circle ($r = 1$), the point on the terminal arm is simply $(\cos\theta, \sin\theta)$.

Using a calculator in radian mode

Put your calculator in radian mode (often shown as RAD). Then $\sin 1 \approx 0.841$. In degree mode the same keys give $\sin 1^\circ \approx 0.0175$, so a wrong mode gives a wrong answer with no warning.

Calculators have no csc, sec, or cot keys. Use the reciprocal: $\csc 2 = \dfrac{1}{\sin 2}$. Don’t use the $\sin^{-1}$ key for this. That key is the inverse sine (it finds an angle), not the reciprocal.

Exact values for the special angles

These come from the special triangles, now with the angles in radians:

$\theta$	$0$	$\tfrac{\pi}{6}$	$\tfrac{\pi}{4}$	$\tfrac{\pi}{3}$	$\tfrac{\pi}{2}$
$\sin\theta$	$0$	$\tfrac{1}{2}$	$\tfrac{\sqrt{2}}{2}$	$\tfrac{\sqrt{3}}{2}$	$1$
$\cos\theta$	$1$	$\tfrac{\sqrt{3}}{2}$	$\tfrac{\sqrt{2}}{2}$	$\tfrac{1}{2}$	$0$
$\tan\theta$	$0$	$\tfrac{\sqrt{3}}{3}$	$1$	$\sqrt{3}$	undefined
$\csc\theta$	undefined	$2$	$\sqrt{2}$	$\tfrac{2\sqrt{3}}{3}$	$1$
$\sec\theta$	$1$	$\tfrac{2\sqrt{3}}{3}$	$\sqrt{2}$	$2$	undefined
$\cot\theta$	undefined	$\sqrt{3}$	$1$	$\tfrac{\sqrt{3}}{3}$	$0$

You don’t need to memorize the bottom three rows: flip the value above. For example, $\sec\tfrac{\pi}{6} = \dfrac{1}{\sqrt{3}/2} = \dfrac{2}{\sqrt{3}} = \dfrac{2\sqrt{3}}{3}$. A ratio is undefined when you’d divide by $0$.

The quadrantal angles $0$, $\tfrac{\pi}{2}$, $\pi$, $\tfrac{3\pi}{2}$, and $2\pi$ sit on the axes. Read their values from the unit circle points $(1, 0)$, $(0, 1)$, $(-1, 0)$, $(0, -1)$, and $(1, 0)$. For example, $\cos\pi = -1$ and $\sin\tfrac{3\pi}{2} = -1$.

The CAST rule and related acute angles

The CAST rule works exactly as before. Only the boundaries are written differently:

Quadrant	Angles	Positive ratios	Angle with related acute angle $\beta$
I	$0$ to $\tfrac{\pi}{2}$	All	$\beta$
II	$\tfrac{\pi}{2}$ to $\pi$	Sine (and csc)	$\pi - \beta$
III	$\pi$ to $\tfrac{3\pi}{2}$	Tangent (and cot)	$\pi + \beta$
IV	$\tfrac{3\pi}{2}$ to $2\pi$	Cosine (and sec)	$2\pi - \beta$

The related acute angle (reference angle) $\beta$ is the angle between the terminal arm and the $x$-axis. The ratio of $\theta$ has the same size as the ratio of $\beta$; CAST gives the sign.

All four angles have related acute angle $\tfrac{\pi}{6}$, so their points differ only in sign.

A quick way to spot the related angle for a multiple of a special angle: look at the denominator. Any multiple of $\tfrac{\pi}{6}$ that isn’t a multiple of $\tfrac{\pi}{2}$ (like $\tfrac{5\pi}{6}$, $\tfrac{7\pi}{6}$, $\tfrac{11\pi}{6}$) has related angle $\tfrac{\pi}{6}$; a multiple of $\tfrac{\pi}{4}$ that isn’t a multiple of $\tfrac{\pi}{2}$ (like $\tfrac{3\pi}{4}$, $\tfrac{5\pi}{4}$) has related angle $\tfrac{\pi}{4}$; a multiple of $\tfrac{\pi}{3}$ that isn’t a multiple of $\pi$ (like $\tfrac{2\pi}{3}$, $\tfrac{4\pi}{3}$) has related angle $\tfrac{\pi}{3}$.

To find the quadrant, compare with $\tfrac{\pi}{2}$, $\pi$, and $\tfrac{3\pi}{2}$. For example, $\tfrac{5\pi}{4}$ is more than $\tfrac{4\pi}{4} = \pi$ but less than $\tfrac{6\pi}{4} = \tfrac{3\pi}{2}$, so it’s in quadrant III.

Worked examples

Example 1: Using a calculator

Evaluate to $3$ decimal places: (a) $\sin 2.5$ (b) $\cos\dfrac{4\pi}{7}$ (c) $\sec 1.2$ (d) $\cot 5$

Solution. In radian mode:

(a) $\sin 2.5 \approx 0.598$

(b) $\cos\dfrac{4\pi}{7} \approx -0.223$

(c) $\sec 1.2 = \dfrac{1}{\cos 1.2} \approx 2.760$

(d) $\cot 5 = \dfrac{1}{\tan 5} \approx -0.296$

Check the signs with CAST: $2.5$ is between $\tfrac{\pi}{2} \approx 1.57$ and $\pi \approx 3.14$, so it’s in quadrant II, where sine is positive. ✓ And $5$ is between $\tfrac{3\pi}{2} \approx 4.71$ and $2\pi \approx 6.28$, so it’s in quadrant IV, where tangent (and so cotangent) is negative. ✓

Example 2: Exact values of primary ratios

Find the exact value of (a) $\sin\dfrac{5\pi}{6}$ (b) $\cos\dfrac{5\pi}{4}$ (c) $\tan\dfrac{5\pi}{3}$

Solution.

(a) $\tfrac{5\pi}{6} = \pi - \tfrac{\pi}{6}$: quadrant II, related angle $\tfrac{\pi}{6}$. Sine is positive in quadrant II.

$$\sin\frac{5\pi}{6} = \sin\frac{\pi}{6} = \frac{1}{2}$$

(b) $\tfrac{5\pi}{4} = \pi + \tfrac{\pi}{4}$: quadrant III, related angle $\tfrac{\pi}{4}$. Cosine is negative in quadrant III.

$$\cos\frac{5\pi}{4} = -\cos\frac{\pi}{4} = -\frac{\sqrt{2}}{2}$$

(c) $\tfrac{5\pi}{3} = 2\pi - \tfrac{\pi}{3}$: quadrant IV, related angle $\tfrac{\pi}{3}$. Tangent is negative in quadrant IV.

$$\tan\frac{5\pi}{3} = -\tan\frac{\pi}{3} = -\sqrt{3}$$

Check (b) on a calculator: $\cos\tfrac{5\pi}{4} \approx -0.7071$ and $-\tfrac{\sqrt{2}}{2} \approx -0.7071$. ✓

Example 3: Exact values of reciprocal ratios

Find the exact value of (a) $\csc\dfrac{7\pi}{6}$ (b) $\sec\dfrac{3\pi}{4}$ (c) $\cot\dfrac{11\pi}{6}$ (d) $\cot\dfrac{3\pi}{2}$

Solution. Find the primary ratio first, then flip it.

(a) $\tfrac{7\pi}{6}$ is in quadrant III with related angle $\tfrac{\pi}{6}$, so $\sin\tfrac{7\pi}{6} = -\tfrac{1}{2}$ and

$$\csc\frac{7\pi}{6} = \frac{1}{-1/2} = -2$$

(b) $\tfrac{3\pi}{4}$ is in quadrant II with related angle $\tfrac{\pi}{4}$, so $\cos\tfrac{3\pi}{4} = -\tfrac{\sqrt{2}}{2}$ and

$$\sec\frac{3\pi}{4} = \frac{1}{-\sqrt{2}/2} = -\frac{2}{\sqrt{2}} = -\sqrt{2}$$

(c) $\tfrac{11\pi}{6}$ is in quadrant IV with related angle $\tfrac{\pi}{6}$, so $\tan\tfrac{11\pi}{6} = -\tfrac{\sqrt{3}}{3} = -\tfrac{1}{\sqrt{3}}$ and

$$\cot\frac{11\pi}{6} = -\sqrt{3}$$

(d) $\tfrac{3\pi}{2}$ is on the negative $y$-axis, at the point $(0, -1)$. Here $\tan\tfrac{3\pi}{2}$ is undefined, so you can’t flip it. Use $\cot\theta = \dfrac{x}{y}$ instead:

$$\cot\frac{3\pi}{2} = \frac{0}{-1} = 0$$

Example 4: From a point on the terminal arm

The point $P(-5, 12)$ is on the terminal arm of an angle $\theta$ in standard position, with $0 \le \theta \le 2\pi$. Find the six trig ratios of $\theta$ exactly, and find $\theta$ in radians to $2$ decimal places.

Solution. $r = \sqrt{(-5)^2 + 12^2} = \sqrt{169} = 13$, so

$$\sin\theta = \frac{12}{13} \qquad \cos\theta = -\frac{5}{13} \qquad \tan\theta = -\frac{12}{5}$$ $$\csc\theta = \frac{13}{12} \qquad \sec\theta = -\frac{13}{5} \qquad \cot\theta = -\frac{5}{12}$$

$P$ is in quadrant II. The related acute angle comes from the side lengths $5$ and $12$, with the calculator in radian mode:

$$\beta = \tan^{-1}\frac{12}{5} \approx 1.176$$

In quadrant II, $\theta = \pi - \beta \approx 3.1416 - 1.1760 \approx 1.97$ rad.

Check: $1.97$ is between $\tfrac{\pi}{2} \approx 1.57$ and $\pi \approx 3.14$, so it’s in quadrant II. ✓ And $\cos 1.9656 \approx -0.385 \approx -\tfrac{5}{13}$. ✓

Common mistakes

Calculator in degree mode. If $\sin\tfrac{\pi}{6}$ doesn’t give exactly $0.5$, you’re in the wrong mode. That’s a quick test you can do any time.

Using sin⁻¹ for cosecant. $\csc 2$ means $\dfrac{1}{\sin 2} \approx 1.100$. The $\sin^{-1}$ key gives an angle, and $\sin^{-1} 2$ is an error, since no sine is bigger than $1$.

Wrong quadrant for angles in radians. Rewrite the boundaries with the same denominator before comparing. Is $\tfrac{7\pi}{4}$ past $\tfrac{3\pi}{2}$? Write $\tfrac{3\pi}{2} = \tfrac{6\pi}{4}$: yes, so it’s in quadrant IV.

Taking the related angle from the y-axis. The related acute angle is always measured to the $x$-axis. For $\tfrac{2\pi}{3}$ it’s $\pi - \tfrac{2\pi}{3} = \tfrac{\pi}{3}$, not $\tfrac{2\pi}{3} - \tfrac{\pi}{2} = \tfrac{\pi}{6}$.

Forgetting the sign. The related angle gives the size of the ratio. CAST gives the sign, and it’s easy to drop: $\cos\tfrac{5\pi}{6} = -\tfrac{\sqrt{3}}{2}$, not $\tfrac{\sqrt{3}}{2}$.

Treating an undefined ratio as 0. $\tan\tfrac{\pi}{2}$ and $\sec\tfrac{\pi}{2}$ are undefined (division by $0$), but $\cot\tfrac{\pi}{2} = 0$. Go back to $x$, $y$, and $r$ when unsure.

Practice

1. (Warm-up) Give the exact value: (a) $\sin\dfrac{\pi}{3}$ (b) $\cos\dfrac{\pi}{4}$ (c) $\tan\dfrac{\pi}{6}$

Solution

(a) $\dfrac{\sqrt{3}}{2}$ (b) $\dfrac{\sqrt{2}}{2}$ (c) $\dfrac{\sqrt{3}}{3}$, which is the same as $\dfrac{1}{\sqrt{3}}$

2. (Warm-up) In which quadrant does each angle’s terminal arm lie? (a) $\dfrac{4\pi}{5}$ (b) $\dfrac{5\pi}{3}$ (c) $4$ (d) $6$

Solution

(a) $\tfrac{4\pi}{5}$ is between $\tfrac{\pi}{2}$ and $\pi$: quadrant II.

(b) $\tfrac{5\pi}{3}$ is between $\tfrac{3\pi}{2} = \tfrac{4.5\pi}{3}$ and $2\pi = \tfrac{6\pi}{3}$: quadrant IV.

(c) $\pi \approx 3.14 \lt 4 \lt \tfrac{3\pi}{2} \approx 4.71$: quadrant III.

(d) $\tfrac{3\pi}{2} \approx 4.71 \lt 6 \lt 2\pi \approx 6.28$: quadrant IV.

3. (Warm-up) Evaluate to $3$ decimal places: (a) $\cos 0.6$ (b) $\tan\dfrac{5\pi}{8}$ (c) $\csc 4$

Solution

(a) $\cos 0.6 \approx 0.825$

(b) $\tan\dfrac{5\pi}{8} \approx -2.414$

(c) $\csc 4 = \dfrac{1}{\sin 4} \approx -1.321$

4. (Core) Find the exact value: (a) $\sin\dfrac{4\pi}{3}$ (b) $\cos\dfrac{7\pi}{4}$ (c) $\tan\dfrac{5\pi}{6}$ (d) $\sin\dfrac{3\pi}{2}$

Solution

(a) Quadrant III, related angle $\tfrac{\pi}{3}$, sine negative: $\sin\dfrac{4\pi}{3} = -\dfrac{\sqrt{3}}{2}$.

(b) Quadrant IV, related angle $\tfrac{\pi}{4}$, cosine positive: $\cos\dfrac{7\pi}{4} = \dfrac{\sqrt{2}}{2}$.

(c) Quadrant II, related angle $\tfrac{\pi}{6}$, tangent negative: $\tan\dfrac{5\pi}{6} = -\dfrac{\sqrt{3}}{3}$.

(d) The point $(0, -1)$: $\sin\dfrac{3\pi}{2} = -1$.

5. (Core) Find the exact value, or say it’s undefined: (a) $\csc\dfrac{5\pi}{4}$ (b) $\sec\dfrac{5\pi}{3}$ (c) $\cot\dfrac{2\pi}{3}$ (d) $\csc\pi$

Solution

(a) $\sin\tfrac{5\pi}{4} = -\tfrac{\sqrt{2}}{2}$, so $\csc\dfrac{5\pi}{4} = -\dfrac{2}{\sqrt{2}} = -\sqrt{2}$.

(b) $\cos\tfrac{5\pi}{3} = \tfrac{1}{2}$, so $\sec\dfrac{5\pi}{3} = 2$.

(c) $\tan\tfrac{2\pi}{3} = -\sqrt{3}$, so $\cot\dfrac{2\pi}{3} = -\dfrac{1}{\sqrt{3}} = -\dfrac{\sqrt{3}}{3}$.

(d) $\sin\pi = 0$, so $\csc\pi = \tfrac{1}{0}$ is undefined.

6. (Core) Find all angles $\theta$ with $0 \le \theta \le 2\pi$ such that $\sin\theta = -\dfrac{\sqrt{3}}{2}$.

Solution

The related acute angle is $\tfrac{\pi}{3}$, since $\sin\tfrac{\pi}{3} = \tfrac{\sqrt{3}}{2}$. Sine is negative in quadrants III and IV:

$$\theta = \pi + \frac{\pi}{3} = \frac{4\pi}{3} \qquad\text{or}\qquad \theta = 2\pi - \frac{\pi}{3} = \frac{5\pi}{3}$$

7. (Core) Evaluate $\cos^2\dfrac{\pi}{6} - \sin^2\dfrac{\pi}{6}$ exactly, and compare it with $\cos\dfrac{\pi}{3}$.

Solution

$$\cos^2\frac{\pi}{6} - \sin^2\frac{\pi}{6} = \left(\frac{\sqrt{3}}{2}\right)^2 - \left(\frac{1}{2}\right)^2 = \frac{3}{4} - \frac{1}{4} = \frac{1}{2}$$

This equals $\cos\tfrac{\pi}{3} = \tfrac{1}{2}$. That’s no accident: it’s an example of a double angle formula you’ll meet in the next unit.

8. (Challenge) An angle $\theta$ has its terminal arm in quadrant III, and $\tan\theta = \dfrac{7}{24}$. Find the exact values of the other five trig ratios, and find $\theta$ in radians, with $0 \le \theta \le 2\pi$, to $2$ decimal places.

Solution

In quadrant III, $x$ and $y$ are both negative, so take the point $(-24, -7)$. Then $r = \sqrt{24^2 + 7^2} = \sqrt{625} = 25$:

$$\sin\theta = -\frac{7}{25} \qquad \cos\theta = -\frac{24}{25} \qquad \csc\theta = -\frac{25}{7} \qquad \sec\theta = -\frac{25}{24} \qquad \cot\theta = \frac{24}{7}$$

The related acute angle is $\beta = \tan^{-1}\tfrac{7}{24} \approx 0.284$. In quadrant III, $\theta = \pi + \beta \approx 3.43$ rad.

9. (Challenge) Find all angles $\theta$ with $0 \le \theta \le 2\pi$ such that (a) $\sec\theta = -\sqrt{2}$ (b) $\cot\theta = -1$

Solution

(a) $\sec\theta = -\sqrt{2}$ means $\cos\theta = -\tfrac{1}{\sqrt{2}} = -\tfrac{\sqrt{2}}{2}$. The related angle is $\tfrac{\pi}{4}$, and cosine is negative in quadrants II and III:

$$\theta = \frac{3\pi}{4} \qquad\text{or}\qquad \theta = \frac{5\pi}{4}$$

(b) $\cot\theta = -1$ means $\tan\theta = -1$. The related angle is $\tfrac{\pi}{4}$, and tangent is negative in quadrants II and IV:

$$\theta = \frac{3\pi}{4} \qquad\text{or}\qquad \theta = \frac{7\pi}{4}$$