Precalculus · F.TF.A.2

Signs of the trig functions by quadrant

You are told only that \cos\theta is negative and \tan\theta is negative. Where on the circle does the angle end, and what does that fix about the other four values?

You need

Trig functions on the unit circlePrecalculus

Unlocks

Power reducing and half angle formulasPrecalculus

Where you are 8 in The unit circle

Trigonometry7

The unit circle8

▴ 2 above Reference angles Secant, cosecant and cotangent on the unit circle

Signs of the trig functions by quadrant

Sine, cosine and tangent as functions of a real number Unit circle symmetry and even-odd identities ▾ Show all 8 (1 more)

When you reach for it

On the unit circle, cosine and sine are coordinates: if the terminal point of \theta is (x, y), then \cos\theta = x and \sin\theta = y. Knowing one value is rarely enough. The circle equation x^2 + y^2 = 1 hands back the other coordinate only up to sign, because a vertical or horizontal line meets the circle twice. Something else has to choose.

That something is usually a second sign. A problem says “\displaystyle \cos\theta = -\frac{55}{73} and \tan\theta < 0”, or “\sin\theta < 0 and \tan\theta > 0”. Reach for this move whenever a value comes with a condition like that, whenever you need the sign of a trig value at an angle you cannot evaluate exactly, and whenever an angle is described only by the region it lies in. It turns two signs into a quadrant, and a quadrant into the sign of all six functions.

The procedure

Where each sign comes from

Write the six functions in terms of the terminal point (x, y). The first three come from the unit circle; the other three are their reciprocals, the secant, cosecant and cotangent, which carry over from right triangles to every angle the same way:

\begin{aligned} \sin\theta &= y & \csc\theta &= \frac{1}{y} \\ \cos\theta &= x & \sec\theta &= \frac{1}{x} \\ \tan\theta &= \frac{y}{x} & \cot\theta &= \frac{x}{y} \end{aligned}

A number and its reciprocal always have the same sign, since their product is 1, which is positive. So the six functions fall into three pairs that share a sign:

sine and cosecant have the sign of y: positive above the x-axis, negative below;
cosine and secant have the sign of x: positive right of the y-axis, negative left;
tangent and cotangent have the sign of \displaystyle \frac{y}{x}: positive when x and y agree in sign (Quadrants I and III), negative when they differ (Quadrants II and IV).

The last rule is the sign rule for dividing signed numbers: a quotient is positive exactly when the two numbers have the same sign. Put together, quadrant by quadrant:

Quadrant	x	y	positive	negative
I	+	+	all six	none
II	-	+	\sin, \csc	\cos, \sec, \tan, \cot
III	-	-	\tan, \cot	\sin, \csc, \cos, \sec
IV	+	-	\cos, \sec	\sin, \csc, \tan, \cot

Figure 1The signs of the terminal point’s coordinates decide every sign. In each quadrant, the functions named are positive and the rest are negative.

There is nothing to memorize beyond “sine is y, cosine is x”. Every row of the table is read off the picture.

From two signs to a quadrant

Each pair cuts the plane in half. The sign of sine (or cosecant) picks the upper or lower half; the sign of cosine (or secant) picks the right or left half; the sign of tangent (or cotangent) picks the diagonal pair I and III or the pair II and IV. Two halves from different pairs overlap in exactly one quadrant, so two signs from different pairs name the quadrant. Two signs from the same pair say the same thing twice and name nothing.

For \cos\theta < 0 and \tan\theta < 0: a negative cosine allows II or III, a negative tangent allows II or IV, and the only quadrant on both lists is II. Then the sign of every other function follows from the table: in Quadrant II, \sin\theta and \csc\theta are positive.

From a quadrant to exact values

The size of an unknown coordinate comes from x^2 + y^2 = 1, or from the distance \displaystyle r = \sqrt{x^2 + y^2} when a point off the circle is known. The quadrant then supplies the sign that the square root left open. The two jobs are separate: algebra gives the size, the quadrant gives the sign.

Worked examples

Example 1

A cosine and the sign of the tangent

\displaystyle \cos\theta = -\frac{55}{73} and \tan\theta < 0. Name the quadrant of \theta and find the other five values.

Solution

A negative cosine puts the terminal point left of the y-axis: Quadrant II or III. A negative tangent means x and y have opposite signs: Quadrant II or IV. Only Quadrant II is on both lists, so \theta is in Quadrant II.

The sine comes from the circle equation:

\begin{aligned} \sin^2\theta &= 1 - \frac{3025}{5329} = \frac{2304}{5329} \\ \sin\theta &= \pm\frac{48}{73} \end{aligned}

The vertical line \displaystyle x = -\frac{55}{73} meets the circle twice, once in each candidate quadrant. A Quadrant II point is above the x-axis, so \displaystyle \sin\theta = \frac{48}{73}.

Figure 2A cosine of \displaystyle -\frac{55}{73} allows two terminal points. The sign of the tangent keeps one of them.

The rest are quotients and reciprocals:

\begin{aligned} \tan\theta &= \frac{48/73}{-55/73} = -\frac{48}{55} \\ \csc\theta &= \frac{73}{48} \qquad \sec\theta = -\frac{73}{55} \\ \cot\theta &= -\frac{55}{48} \end{aligned}

Check. The tangent came out negative, as given, and only \sin and \csc are positive, which is the Quadrant II row of the table.

Example 2

Naming quadrants from pairs of signs

Name the quadrant of \theta, or say why it cannot be named.

(a) \sin\theta < 0 and \tan\theta > 0 \qquad (b) \sec\theta > 0 and \csc\theta < 0

Solution

(a) y < 0 allows III or IV. A positive tangent allows I or III. Quadrant III.

(b) The secant has the sign of x, so x > 0: I or IV. The cosecant has the sign of y, so y < 0: III or IV. Quadrant IV.

(d) Both conditions say y > 0, because cosecant always has the sign of sine. The angle is in Quadrant I or Quadrant II, and nothing given decides between them. A second piece of information about x, or about the tangent, is needed.

Example 3

Signs without evaluating anything

Decide the sign of each value without a calculator.

(a) \cos 250^\circ \quad (b) \tan(-100^\circ) \quad (c) \displaystyle \csc\frac{5\pi}{4} \quad (d) \displaystyle \sec\frac{11\pi}{6} \quad (e) \sin 2

Solution

Only the quadrant matters, so place each angle first.

(a) 180^\circ < 250^\circ < 270^\circ: Quadrant III, where x < 0. Negative.

(b) -100^\circ + 360^\circ = 260^\circ, a coterminal angle in Quadrant III, so -100^\circ ends there too. Tangent is positive in III. Positive.

(c) \displaystyle \frac{5\pi}{4} is \pi plus a quarter of \pi: Quadrant III, where y < 0. Cosecant has the sign of y. Negative.

(d) \displaystyle \frac{11\pi}{6} is \displaystyle \frac{\pi}{6} short of 2\pi: Quadrant IV, where x > 0. Positive.

(e) The 2 is in radians, and \displaystyle \frac{\pi}{2} \approx 1.571 < 2 < \pi \approx 3.142: Quadrant II, where y > 0. Positive.

The size of each value is a separate question, answered with reference angles; the sign needed only the quadrant.

Example 4

A tangent and the sign of the sine

\displaystyle \tan\theta = \frac{44}{117} and \sin\theta < 0. Find the other five values.

Solution

A positive tangent allows I or III; a negative sine allows III or IV. So \theta is in Quadrant III, where both coordinates are negative.

The tangent is the slope of the terminal side, so the side lies on the line through the origin with slope \displaystyle \frac{44}{117}. The Quadrant III ray of that line passes through (-117, -44), at distance

\begin{aligned} r &= \sqrt{117^2 + 44^2} = \sqrt{13{,}689 + 1936} \\ &= \sqrt{15{,}625} = 125 \end{aligned}

Dividing by r gives the terminal point, so

\begin{aligned} \sin\theta &= -\frac{44}{125} \qquad \csc\theta = -\frac{125}{44} \\ \cos\theta &= -\frac{117}{125} \qquad \sec\theta = -\frac{125}{117} \\ \cot\theta &= \frac{117}{44} \end{aligned}

Choosing (117, 44) instead would have given the Quadrant I ray of the same line: the right tangent, the wrong signs. The quadrant is what picks the ray.

Example 5

A cosecant and the sign of the cotangent

\displaystyle \csc\theta = -\frac{157}{85} and \cot\theta < 0. Find the other five values.

Solution

The sine is the reciprocal of the cosecant: \displaystyle \sin\theta = -\frac{85}{157}, so y < 0 and \theta is in III or IV. A negative cotangent allows II or IV. Quadrant IV, so x > 0.

\begin{aligned} \cos^2\theta &= 1 - \frac{7225}{24{,}649} \\ &= \frac{17{,}424}{24{,}649} \\ \cos\theta &= \frac{132}{157} \end{aligned}

taking the positive root because x > 0 in Quadrant IV. Then

\begin{aligned} \tan\theta &= -\frac{85}{132} \qquad \sec\theta = \frac{157}{132} \\ \cot\theta &= -\frac{132}{85} \end{aligned}

Only \cos and \sec came out positive, the Quadrant IV pattern.

Example 6

The tip of a compass needle

Paora lays a compass on grid paper with its center at the origin, east along the positive x-axis and north along the positive y-axis. He traces the line of the needle and finds its slope is \displaystyle \frac{12}{5}. The painted tip of the needle is south of the center, and the needle reaches 6.5 cm from the center to the tip. Let \theta be the angle in standard position whose terminal side runs through the tip. Find \cos\theta, \sin\theta and the position of the tip.

Solution

The line through the center has slope \displaystyle \tan\theta = \frac{12}{5} > 0, so the tip is in I or III. South of the center means y < 0: III or IV. The tip is in Quadrant III, southwest of the center.

The Quadrant III ray of the line passes through (-5, -12), at distance \displaystyle \sqrt{25 + 144} = 13, so

\cos\theta = -\frac{5}{13} \qquad \sin\theta = -\frac{12}{13}

The tip is 6.5 cm out along this ray, so it is the terminal point scaled by 6.5:

\begin{aligned} x &= 6.5 \cdot \left(-\frac{5}{13}\right) = -2.5 \\ y &= 6.5 \cdot \left(-\frac{12}{13}\right) = -6 \end{aligned}

The tip is 2.5 cm west and 6 cm south of the center. The slope alone fixed only the line of the needle; the word “south” chose which end of it.

Figure 3The painted tip in Example 6 lies 2.5 cm west and 6 cm south of the center.

Example 7

Halving and doubling an angle

An angle satisfies 180^\circ < \theta < 270^\circ. Find the signs of \displaystyle \sin\frac{\theta}{2}, \displaystyle \cos\frac{\theta}{2} and \displaystyle \tan\frac{\theta}{2}, and decide whether \sin 2\theta and \cos 2\theta have a fixed sign.

Solution

Divide the inequality by 2: \displaystyle 90^\circ < \frac{\theta}{2} < 135^\circ. That is inside Quadrant II, so \displaystyle \sin\frac{\theta}{2} > 0, \displaystyle \cos\frac{\theta}{2} < 0 and \displaystyle \tan\frac{\theta}{2} < 0.

Multiply it by 2 instead: 360^\circ < 2\theta < 540^\circ. Subtracting a full turn gives the coterminal range 0^\circ to 180^\circ, so 2\theta ends in Quadrant I, in Quadrant II, or on the positive y-axis (when \theta = 225^\circ). In all three, y > 0, so \sin 2\theta > 0. The cosine has no fixed sign: 2\theta is in Quadrant I for \theta < 225^\circ, where \cos 2\theta > 0, and in Quadrant II for \theta > 225^\circ, where \cos 2\theta < 0.

The answer for \sin 2\theta agrees with a formula met later, \sin 2\theta = 2\sin\theta\cos\theta: in Quadrant III both \sin\theta and \cos\theta are negative, and their product is positive. Choosing the sign of a half angle in exactly this way is the last step of the half angle formulas in Power reducing and half angle formulas.

Boundaries

A zero is not a sign. When one of the given values is 0, the terminal side lies on an axis and \theta is in no quadrant at all. At such angles some of the six functions are undefined; see Exact values on the unit circle and Secant, cosecant and cotangent on the unit circle. Every condition in this lesson is a strict inequality for that reason.

The two signs must come from different pairs. Sine with cosecant, cosine with secant, or tangent with cotangent repeat one fact and leave two quadrants open, as in Example 2(d).

Some pairs of conditions are impossible. \tan\theta > 0 with \cot\theta < 0, or \sin\theta > 0 with \csc\theta < 0, describe no angle, since a reciprocal keeps its sign.

The quadrant fixes the sign, not the size. Knowing \theta is in Quadrant II says \sin\theta > 0 but nothing about whether it is 0.1 or 0.9.

A quadrant is a set of terminal sides, not of angles between two numbers. -100^\circ, 260^\circ and 620^\circ all end in Quadrant III. Bring an angle inside one turn before reading its quadrant.

Where this goes next

The quadrant supplies the sign; reference angles supply the size, and together they give any value from an acute one. Reflecting a terminal point across an axis changes exactly the signs this lesson tracks, which is Unit circle symmetry and even-odd identities. The same “size from the algebra, sign from the quadrant” choice comes back whenever a square root is taken in trigonometry: in the Pythagorean identity solved for one function, in half angle formulas, and in finding a direction angle from the components of a vector, where Magnitude and direction of a vector gives the length and the signs of the components give the quadrant.

Exercises

Naming the quadrant

Name the quadrant of \theta.

\sin\theta < 0 and \cos\theta < 0
\tan\theta < 0 and \sin\theta < 0
\sec\theta < 0 and \cot\theta > 0
\csc\theta > 0 and \sec\theta < 0
\cos\theta > 0 and \cot\theta > 0
\cot\theta < 0 and \cos\theta < 0

Signs without a calculator

Decide whether each value is positive or negative.

\sin 305^\circ
\cot 165^\circ
\sec(-130^\circ)
\displaystyle \tan\frac{11\pi}{8}
\csc 4

All six from one value and one sign

Find the other five values exactly, with rational denominators.

\displaystyle \sin\theta = -\frac{7}{9} and \cos\theta < 0
\displaystyle \cos\theta = \frac{65}{97} and \sin\theta < 0
\displaystyle \tan\theta = -\frac{60}{91} and \sec\theta > 0
\sec\theta = -3 and \tan\theta > 0
\cot\theta = -2 and \csc\theta > 0
\sin\theta = 0.6 and \tan\theta < 0

Compass needles

Each compass sits on grid paper with its center at the origin, east along the positive x-axis and north along the positive y-axis; \theta is the angle in standard position whose terminal side runs through the painted tip.

Ihaia’s needle reaches 6.5 cm from the center to the painted tip. The line of the needle has slope \displaystyle -\frac{56}{33}, and the tip is east of the center. Find \cos\theta, \sin\theta and the position of the tip.
Sizwe’s needle reaches 4 cm from the center to the tip, \cos\theta = -0.25, and the tip is north of the center. Find \sin\theta exactly and the position of the tip to the nearest hundredth of a centimeter.

Reasoning

Explain why \sec\theta always has the same sign as \cos\theta, and name the other two pairs of functions that always share a sign.
Is there an angle with \tan\theta > 0 and \cot\theta < 0? Explain.
Explain why the signs of \sec\theta and \cos\theta together never name a quadrant, but the signs of \sec\theta and \csc\theta always do.
Explain why knowing the signs of \cos\theta and \tan\theta fixes the sign of \sin\theta.
In which quadrants is \sin\theta\cos\theta < 0? Explain.
\theta is in Quadrant II. In which quadrant is \theta + 180^\circ, and how do the signs of the six functions there compare with their signs at \theta?

Answers

Quadrant III. y < 0 gives III or IV; x < 0 gives II or III.
Quadrant IV. A negative tangent gives II or IV; y < 0 gives III or IV.
Quadrant III. \sec\theta < 0 means x < 0: II or III. A positive cotangent gives I or III.
Quadrant II. \csc\theta > 0 means y > 0: I or II. \sec\theta < 0 means x < 0: II or III.
Quadrant I. x > 0 gives I or IV; a positive cotangent gives I or III.
Quadrant II. A negative cotangent gives II or IV; x < 0 gives II or III.
Negative. 305^\circ is in Quadrant IV, where y < 0.
Negative. 165^\circ is in Quadrant II, where x < 0 < y, so \displaystyle \frac{x}{y} < 0.
Negative. -130^\circ is coterminal with 230^\circ, in Quadrant III, where x < 0.
Positive. \displaystyle \pi < \frac{11\pi}{8} < \frac{3\pi}{2}, Quadrant III, where tangent is positive.
Negative. \displaystyle \pi \approx 3.142 < 4 < \frac{3\pi}{2} \approx 4.712, Quadrant III, where y < 0.
Quadrant III. \displaystyle \cos\theta = -\frac{4\sqrt{2}}{9}, \displaystyle \tan\theta = \frac{7\sqrt{2}}{8}, \displaystyle \csc\theta = -\frac{9}{7}, \displaystyle \sec\theta = -\frac{9\sqrt{2}}{8}, \displaystyle \cot\theta = \frac{4\sqrt{2}}{7}. \displaystyle \cos^2\theta = 1 - \frac{49}{81} = \frac{32}{81}.
Quadrant IV. \displaystyle \sin\theta = -\frac{72}{97}, \displaystyle \tan\theta = -\frac{72}{65}, \displaystyle \csc\theta = -\frac{97}{72}, \displaystyle \sec\theta = \frac{97}{65}, \displaystyle \cot\theta = -\frac{65}{72}. \displaystyle \sin^2\theta = 1 - \frac{4225}{9409} = \frac{5184}{9409}.
Quadrant IV. \displaystyle \sin\theta = -\frac{60}{109}, \displaystyle \cos\theta = \frac{91}{109}, \displaystyle \csc\theta = -\frac{109}{60}, \displaystyle \sec\theta = \frac{109}{91}, \displaystyle \cot\theta = -\frac{91}{60}. The ray passes through (91, -60) and \displaystyle r = \sqrt{8281 + 3600} = 109.
Quadrant III. \displaystyle \cos\theta = -\frac{1}{3}, \displaystyle \sin\theta = -\frac{2\sqrt{2}}{3}, \displaystyle \tan\theta = 2\sqrt{2}, \displaystyle \csc\theta = -\frac{3\sqrt{2}}{4}, \displaystyle \cot\theta = \frac{\sqrt{2}}{4}. \displaystyle \sin^2\theta = 1 - \frac{1}{9} = \frac{8}{9}.
Quadrant II. \displaystyle \tan\theta = -\frac{1}{2}, \displaystyle \sin\theta = \frac{\sqrt{5}}{5}, \displaystyle \cos\theta = -\frac{2\sqrt{5}}{5}, \displaystyle \csc\theta = \sqrt{5}, \displaystyle \sec\theta = -\frac{\sqrt{5}}{2}. The ray passes through (-2, 1) and \displaystyle r = \sqrt{5}.
Quadrant II. \cos\theta = -0.8, \tan\theta = -0.75, \displaystyle \csc\theta = \frac{5}{3}, \sec\theta = -1.25, \displaystyle \cot\theta = -\frac{4}{3}. \cos^2\theta = 1 - 0.36 = 0.64.
\displaystyle \cos\theta = \frac{33}{65}, \displaystyle \sin\theta = -\frac{56}{65}, tip at (3.3, -5.6): 3.3 cm east and 5.6 cm south. A negative slope with x > 0 is Quadrant IV; the ray passes through (33, -56) with r = 65, and \displaystyle 6.5 \cdot \frac{33}{65} = 3.3.
\displaystyle \sin\theta = \frac{\sqrt{15}}{4}; the tip is at \displaystyle (-1, \sqrt{15}) \approx (-1, 3.87). \displaystyle \sin^2\theta = 1 - \frac{1}{16} = \frac{15}{16}, positive since the tip is north (Quadrant II), and the tip is 4(\cos\theta, \sin\theta).
\displaystyle \sec\theta = \frac{1}{\cos\theta}, and \cos\theta \cdot \sec\theta = 1 > 0, so the two are both positive or both negative. The same holds for \sin\theta and \csc\theta, and for \tan\theta and \cot\theta.
No. \tan\theta \cdot \cot\theta = 1 wherever both are defined, and a product of a positive and a negative number is negative, never 1.
Secant and cosine both have the sign of x, so together they give only a left or right half, two quadrants. Secant gives the sign of x and cosecant the sign of y, and the two signs together are exactly what a quadrant is.
The cosine gives the sign of x, and the tangent \displaystyle \frac{y}{x} is positive exactly when y has the same sign as x. So y, which is \sin\theta, has the sign of x when the tangent is positive and the opposite sign when it is negative.
Quadrants II and IV. \sin\theta\cos\theta = xy, which is negative exactly when x and y have opposite signs. (It is also the pair where \tan\theta < 0, since \displaystyle \frac{y}{x} and xy have the same sign.)
Quadrant IV. A half turn sends the terminal point (x, y) to (-x, -y), so sine, cosecant, cosine and secant all change sign, while \displaystyle \frac{-y}{-x} = \frac{y}{x} keeps tangent and cotangent negative, as in Quadrant II.