Signs of trigonometric ratios and standard values

Once angles can lie in any quad­rant, every trigono­met­ric ratio car­ries a sign, pos­i­tive or neg­a­tive. Know­ing the sign at a glance, together with the exact val­ues at $0°$, $30°$, $45°$, $60°$ and $90°$, lets you find the ratios of any angle in the lessons that fol­low. This les­son explains where the signs come from and sets out the table of stan­dard val­ues.

The four quad­rants#

The axes $X'OX$ and $Y'OY$ divide the plane into four quad­rants, num­bered anti­clock­wise from the pos­i­tive $x$-axis.

A com­mon way to remem­ber the pos­i­tive ratios, going round from the first quad­rant: All, Sin, Tan, Cos.

Why the signs are what they are#

The dis­tance $r = OP$ is always pos­i­tive, so the sign of each ratio depends only on the signs of $x$ and $y$.

Take an angle in the sec­ond quad­rant, $90° < \theta < 180°$. Then $P(x, y)$ lies in the sec­ond quad­rant, where $x < 0$ and $y > 0$. So

$$\displaystyle \sin\theta = \frac{y}{r} > 0, \qquad \cos\theta = \frac{x}{r} < 0, \qquad \tan\theta = \frac{y}{x} < 0$$

and their rec­i­p­ro­cals fol­low: $\csc\theta > 0$, $\sec\theta < 0$, $\cot\theta < 0$.

The same rea­son­ing gives the other quad­rants. In the third quad­rant both $x$ and $y$ are neg­a­tive, so $\displaystyle \tan\theta = \frac{y}{x}$ is pos­i­tive while $\sin$ and $\cos$ are neg­a­tive. In the fourth quad­rant $x > 0$ and $y < 0$, so only $\cos$ and $\sec$ are pos­i­tive.

Exam­ple 1#

State the sign of (a) $\sin 200°$ (b) $\cos 300°$ (c) $\tan 135°$.

(a) $200°$ is in the third quad­rant, where $\sin$ is neg­a­tive.

(b) $300°$ is in the fourth quad­rant, where $\cos$ is pos­i­tive.

(c) $135°$ is in the sec­ond quad­rant, where $\tan$ is neg­a­tive.

Exam­ple 2#

If $\displaystyle \cos\theta = -\frac{12}{13}$ and $\theta$ lies in the third quad­rant, find $\sin\theta$ and $\tan\theta$.

$\displaystyle \sin^2\theta = 1 - \frac{144}{169} = \frac{25}{169}$, so $\displaystyle \sin\theta = \pm\frac{5}{13}$. In the third quad­rant $\sin$ is neg­a­tive, so $\displaystyle \sin\theta = -\frac{5}{13}$. Then

$$\displaystyle \tan\theta = \frac{\sin\theta}{\cos\theta} = \frac{-5/13}{-12/13} = \frac{5}{12}$$

which is pos­i­tive, as it should be in the third quad­rant.

A quick method: draw a right tri­an­gle with sides $5$, $12$, $13$ to get the sizes, then attach the signs from the quad­rant.

The table of stan­dard val­ues#

Some pat­terns help you remem­ber it. The $\sin$ row increases from $0$ to $1$; the $\cos$ row is the $\sin$ row back­wards. Each $\tan$ entry is $\sin$ divided by $\cos$. At $90°$, $\displaystyle \tan 90° = \frac{1}{0}$, which is not defined (it is some­times writ­ten $\infty$ to show it grows with­out limit as $\theta$ approaches $90°$).

The other three ratios come from the rec­i­p­ro­cal rela­tions:

$$\displaystyle \cot 30° = \frac{1}{\tan 30°} = \sqrt{3}, \qquad \csc 60° = \frac{1}{\sin 60°} = \frac{2}{\sqrt{3}}, \qquad \sec 60° = \frac{1}{\cos 60°} = 2$$

Exam­ple 3#

Eval­u­ate $\sin^2 30° + \cos^2 60° + \tan^2 45°$.

$$\displaystyle \left(\frac{1}{2}\right)^2 + \left(\frac{1}{2}\right)^2 + 1^2 = \frac{1}{4} + \frac{1}{4} + 1 = \frac{3}{2}$$

Exam­ple 4#

Eval­u­ate $4\cos^3 60° - 3\cos 60°$.

$$\displaystyle 4\left(\frac{1}{2}\right)^3 - 3\left(\frac{1}{2}\right) = \frac{4}{8} - \frac{3}{2} = \frac{1}{2} - \frac{3}{2} = -1$$

(You will see later that $4\cos^3\theta - 3\cos\theta = \cos 3\theta$, and indeed $\cos 180° = -1$.)

Exam­ple 5#

If $\displaystyle \sin\theta = \frac{1}{2}$ and $\cos\theta < 0$, find $\tan\theta$.

$\sin$ is pos­i­tive and $\cos$ is neg­a­tive, so $\theta$ is in the sec­ond quad­rant. The size of $\cos\theta$ is $\displaystyle \sqrt{1 - \frac{1}{4}} = \frac{\sqrt{3}}{2}$, so $\displaystyle \cos\theta = -\frac{\sqrt{3}}{2}$ and

$$\displaystyle \tan\theta = \frac{1/2}{-\sqrt{3}/2} = -\frac{1}{\sqrt{3}}$$

Prac­tice#

1. State the sign of $\sin 250°$, $\cos 100°$, $\tan 300°$ and $\sec 340°$.
2. In which quad­rant is $\sin\theta > 0$ and $\cos\theta < 0$?
3. If $\displaystyle \sin\theta = -\frac{4}{5}$ and $\theta$ is in the fourth quad­rant, find $\cos\theta$ and $\tan\theta$.
4. If $\displaystyle \tan\theta = -\frac{5}{12}$ and $\theta$ is in the sec­ond quad­rant, find $\sin\theta$ and $\cos\theta$.
5. Find $\sec 45°$, $\csc 30°$ and $\cot 60°$.
6. Eval­u­ate $\sin 30° \cos 60° + \cos 30° \sin 60°$.
7. Eval­u­ate $\tan^2 60° + 2\tan^2 45°$.
8. Eval­u­ate $(\sin 45° + \cos 45°)^2$.
9. Show that $2\sin 30° \cos 30° = \sin 60°$.
10. If $\displaystyle \cos\theta = -\frac{\sqrt{3}}{2}$ and $\sin\theta > 0$, find $\tan\theta$.

Answers#