Math Core

Before you compute a trig value, it helps to know whether the answer will be positive or negative. That one piece of information catches sign errors, and it lets you find every trig function of an angle from just one of them. The whole story comes down to the signs of $x$ and $y$ in each quadrant.

Where the signs come from

For an angle $\theta$ in standard position with a point $(x, y)$ on its terminal side,

\sin\theta = \frac{y}{r}, \qquad \cos\theta = \frac{x}{r}, \qquad \tan\theta = \frac{y}{x}.

The distance $r$ is always positive, so it never affects a sign. That means:

$\sin\theta$ has the same sign as $y$ .
$\cos\theta$ has the same sign as $x$ .
$\tan\theta$ is positive when $x$ and $y$ have the same sign, and negative when they have opposite signs.

The reciprocal functions copy their partners, because $\dfrac{1}{\text{positive}}$ is positive and $\dfrac{1}{\text{negative}}$ is negative. So $\csc\theta$ matches $\sin\theta$ , $\sec\theta$ matches $\cos\theta$ , and $\cot\theta$ matches $\tan\theta$ .

Now go around the quadrants:

quadrant	signs of $(x, y)$	$\sin\theta$ , $\csc\theta$	$\cos\theta$ , $\sec\theta$	$\tan\theta$ , $\cot\theta$
I	$(+, +)$	$+$	$+$	$+$
II	$(-, +)$	$+$	$-$	$-$
III	$(-, -)$	$-$	$-$	$+$
IV	$(+, -)$	$-$	$+$	$-$

Which trig functions are positive in each quadrant, with the signs of (x, y).

All Students Take Calculus

Starting in Quadrant I and going counterclockwise, the functions that are positive are:

I: All of them
II: Sine (and cosecant)
III: Tangent (and cotangent)
IV: Cosine (and secant)

Everything else is negative in that quadrant.

If you forget the phrase, don't panic. Just ask yourself whether $x$ and $y$ are positive or negative in that quadrant, and reason from the definitions. That's all the mnemonic is doing.

Worked example: Predicting a sign

Without a calculator, decide whether each value is positive or negative.

$\sin 200^\circ$
$\cos 300^\circ$
$\tan\dfrac{2\pi}{3}$

Solutions.

$200^\circ$ is between $180^\circ$ and $270^\circ$ , in Quadrant III. There $y < 0$ , so $\sin 200^\circ$ is negative.
$300^\circ$ is between $270^\circ$ and $360^\circ$ , in Quadrant IV. There $x > 0$ , so $\cos 300^\circ$ is positive.
$\dfrac{2\pi}{3}$ is between $\dfrac{\pi}{2}$ and $\pi$ , in Quadrant II. There $x < 0$ and $y > 0$ , so $\tan\dfrac{2\pi}{3}$ is negative.

Finding the quadrant from the signs

You can also run the table backwards. Each condition rules out two quadrants, and two conditions together usually leave exactly one.

Worked example: Two conditions, one quadrant

In which quadrant is $\theta$ if $\sin\theta < 0$ and $\tan\theta > 0$ ?

$\sin\theta < 0$ means $y < 0$ : Quadrant III or IV.
$\tan\theta > 0$ means $x$ and $y$ have the same sign: Quadrant I or III.

The only quadrant on both lists is Quadrant III.

Finding all the trig values from one

Here is where signs really pay off. Suppose you know one trig value and which quadrant $\theta$ is in. Then you can find all the others:

Use the known value to pick numbers for two of $x$ , $y$ and $r$ . Keep $r$ positive and give $x$ or $y$ the sign the quadrant requires.
Find the third using $x^2 + y^2 = r^2$ . Choose its sign from the quadrant.
Write each trig function from the definitions.

Worked example: Given cosine and the quadrant

If $\cos\theta = -\dfrac{5}{13}$ and $\theta$ is in Quadrant III, find $\sin\theta$ and $\tan\theta$ .

Cosine is $\dfrac{x}{r}$ , so take $x = -5$ and $r = 13$ . Then

(-5)^2 + y^2 = 13^2 \quad\Rightarrow\quad y^2 = 144 \quad\Rightarrow\quad y = \pm 12.

In Quadrant III, $y$ is negative, so $y = -12$ . Therefore

\sin\theta = \frac{-12}{13} = -\frac{12}{13}, \qquad \tan\theta = \frac{-12}{-5} = \frac{12}{5}.

Check: in Quadrant III, sine should be negative and tangent positive. Both are.

Worked example: Given tangent and a sign condition

If $\tan\theta = -2$ and $\sin\theta > 0$ , find $\sin\theta$ and $\cos\theta$ .

First find the quadrant. $\tan\theta < 0$ puts $\theta$ in Quadrant II or IV, and $\sin\theta > 0$ puts it in Quadrant I or II. So $\theta$ is in Quadrant II, where $x < 0$ and $y > 0$ .

Write $\tan\theta = \dfrac{y}{x} = \dfrac{2}{-1}$ , so take $y = 2$ and $x = -1$ . Then $r = \sqrt{(-1)^2 + 2^2} = \sqrt{5}$ , and

\sin\theta = \frac{2}{\sqrt{5}} = \frac{2\sqrt{5}}{5}, \qquad \cos\theta = \frac{-1}{\sqrt{5}} = -\frac{\sqrt{5}}{5}.

Common mistake

When you solve $x^2 + y^2 = r^2$ for the missing coordinate, the square root gives two answers, like $y = \pm 12$ . Don't automatically take the positive one. The quadrant decides the sign. Likewise, when you split a negative ratio such as $-2$ into $\dfrac{y}{x}$ , put the negative sign on whichever coordinate is negative in that quadrant, never on $r$ .

Tip

After finding all the values, check them against the "All Students Take Calculus" pattern. If you're in Quadrant IV and your sine came out positive, something went wrong.

Practice

Practice 1

In which quadrant is $\theta$ if $\cos\theta > 0$ and $\sin\theta < 0$ ?

Practice 2

Is $\tan 250^\circ$ positive or negative?

Practice 3

Which of these values is negative?

Practice 4

If $\sin\theta = \dfrac{4}{5}$ and $\theta$ is in Quadrant II, find $\cos\theta$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 5

If $\cos\theta = \dfrac{12}{13}$ and $\sin\theta < 0$ , find $\tan\theta$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 6

If $\tan\theta = \dfrac{3}{4}$ and $\theta$ is in Quadrant III, find $\sin\theta$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 7

If $\sec\theta = -3$ and $\tan\theta > 0$ , find the exact value of $\sin\theta$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 8

If $\sin\theta \cos\theta < 0$ , where can the terminal side of $\theta$ be?