Magpie Tuition · Higher Maths

Radians and quadrant signs

Allow 20–25 minutes. No calculator is needed. You will locate angles, choose signs and evaluate exact trigonometric values. This prepares you for solving trig equations later.

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Overview

Big Picture

Degrees and radians measure the same thing—an angle—with different units. A full turn is $360^\circ=2\pi$ radians; half a turn is $180^\circ=\pi$ radians.

One radian is the angle at the centre of a circle when the arc length equals the radius. A full circumference is $2\pi$ radii long, which explains $2\pi$ radians in a full turn.

\text{radians}=\text{degrees}\times\frac{\pi}{180}, \qquad \text{degrees}=\text{radians}\times\frac{180}{\pi}.

An angle and its sine are different quantities. For example, $\pi/2$ is an angle; $\sin(\pi/2)=1$ is a ratio. Removing $\pi$ from an angle does not calculate its sine.

Key angles

Degrees	$0^\circ$	$30^\circ$	$45^\circ$	$60^\circ$	$90^\circ$	$180^\circ$	$270^\circ$	$360^\circ$
Radians	$0$	$\pi/6$	$\pi/4$	$\pi/3$	$\pi/2$	$\pi$	$3\pi/2$	$2\pi$

Count anticlockwise from the positive horizontal axis for positive angles; clockwise for negative angles.

Key idea

The Why — the unit circle

On a circle of radius 1, the point at angle $\theta$ has coordinates $(\cos\theta,\sin\theta)$ . Cosine is the horizontal coordinate, sine the vertical coordinate. Tangent is $\sin\theta/\cos\theta$ , when cosine is non-zero.

Redrawn for this page from the resource's text schematic.

Original schematic; positions are illustrative. Axis points are boundaries, not inside any quadrant.

Quadrant	Radian interval	sin	cos	tan
I	$0<\theta<\pi/2$	+	+	+
II	$\pi/2<\theta<\pi$	+	−	−
III	$\pi<\theta<3\pi/2$	−	−	+
IV	$3\pi/2<\theta<2\pi$	−	+	−

Thus, for example, a negative cosine permits quadrants II and III. It does not tell you one unique angle. For a point on an axis, use its coordinates directly: sine or cosine can be zero. Tangent is undefined at $\pi/2$ and $3\pi/2$ , where you would divide by zero.

Method

Step by Step

Read the unit and interval. In Higher questions, an angle without a degree symbol is taken in radians. Keep all angle arithmetic in one unit.
Locate the quadrant by comparing with $\pi/2,\pi,3\pi/2,2\pi$ .
Find the reference angle $\alpha$ : the acute angle between the terminal ray and the horizontal axis.
Choose the correct exact value, then apply the sign from the quadrant.

For angles strictly inside the four quadrants in one positive turn:

Quadrant	Reference angle $\alpha$	Rebuild $\theta$ from $\alpha$
I	$\theta$	$\alpha$
II	$\pi-\theta$	$\pi-\alpha$
III	$\theta-\pi$	$\pi+\alpha$
IV	$2\pi-\theta$	$2\pi-\alpha$

Reference values

$\alpha$	$\sin\alpha$	$\cos\alpha$	$\tan\alpha$
$\pi/6$	$1/2$	$\sqrt3/2$	$1/\sqrt3=\sqrt3/3$
$\pi/4$	$\sqrt2/2$	$\sqrt2/2$	$1$
$\pi/3$	$\sqrt3/2$	$1/2$	$\sqrt3$

These come from two familiar right triangles. Bisect an equilateral triangle of side 2: its right-triangle sides are $1,\sqrt3,2$ . A right isosceles triangle has sides $1,1,\sqrt2$ . Use opposite/hypotenuse for sine and adjacent/hypotenuse for cosine. The quadrant changes the signs, while the reference triangle supplies the magnitudes.

Worked example

Worked examples

1. Sine in quadrant II

Find $\sin(2\pi/3)$ .

Since $\pi/2<2\pi/3<\pi$ , the angle lies in quadrant II. Its reference angle is

\alpha=\pi-\frac{2\pi}{3}=\frac{\pi}{3}.

The sine magnitude is $\sqrt3/2$ , and the vertical coordinate is positive:

\boxed{\sin(2\pi/3)=\sqrt3/2}.

2. Cosine in quadrant III

Find $\cos(5\pi/4)$ .

$\pi<5\pi/4<3\pi/2$ , so use quadrant III. Then $\alpha=5\pi/4-\pi=\pi/4$ . The cosine magnitude is $\sqrt2/2$ , but the horizontal coordinate is negative:

\boxed{\cos(5\pi/4)=-\sqrt2/2}.

3. Tangent in quadrant IV

Find $\tan(11\pi/6)$ .

The angle lies between $3\pi/2=9\pi/6$ and $2\pi=12\pi/6$ , so it is in quadrant IV. Its reference angle is $12\pi/6-11\pi/6=\pi/6$ .

Here sine is negative and cosine positive:

\tan(11\pi/6)=\frac{-1/2}{\sqrt3/2} =-\frac1{\sqrt3}=\boxed{-\frac{\sqrt3}{3}}.

4. A direction in the requested interval

A ray is $\pi/4$ clockwise from the positive horizontal axis. Give its angle in $0\le\theta<2\pi$ .

The clockwise angle is $-\pi/4$ . Add one full turn to express that same ray in the requested interval:

-\frac\pi4+2\pi=\boxed{\frac{7\pi}{4}}.

This is angle placement, not a method for solving a full trig equation.

Common mistake

The Trap

$2\pi$ radians is a full turn; it is not $2\pi$ degrees. In a radian calculation use $\pi-\alpha$ , not $180-\alpha$ .
A correct reference angle still needs the correct function and sign. At $\pi/3$ , sine and cosine have different values.
$0\le\theta<2\pi$ includes $0$ , excludes $2\pi$ . Those angles point along the same ray, but only one belongs to this interval.
For a calculator check of radian inputs, select RAD mode. Calculator decimals are not a replacement for requested exact values.

An unexpected answer may fit one of these slips. Check the unit, quadrant, reference value and sign separately before deciding what needs revision.

Practice

Your Turn

Practice only — these are not your fresh-check questions. Give exact answers. For trig values, show quadrant, reference angle and sign.

Convert (a) $135^\circ$ , (b) $300^\circ$ to radians; convert (c) $3\pi/4$ , (d) $11\pi/6$ to degrees.
For each angle, state its quadrant and the signs of sin, cos and tan: (a) $\pi/5$ , (b) $4\pi/5$ , (c) $6\pi/5$ , (d) $9\pi/5$ .
For angles not on an axis, which quadrants are possible when (a) sine is negative, (b) cosine is negative, (c) tangent is positive? Which quadrant has both negative sine and negative cosine?
Evaluate (a) $\sin(\pi/3)$ , (b) $\sin(4\pi/3)$ , (c) $\cos(7\pi/4)$ , (d) $\tan(2\pi/3)$ .
Find the reference angle of $5\pi/3$ , then find its sine and cosine.
A downward-pointing ray is labelled $3\pi/2$ . Express its direction in $-\pi\le\theta<\pi$ . For this same interval, which of $-\pi$ and $\pi$ is included?
One answer claims $\sin(2\pi/3)=2/3$ ; another claims it is $1/2$ . Explain a possible slip behind each answer and give the correct value.
Use the axis points to find (a) $\sin(3\pi/2)$ , (b) $\cos\pi$ , (c) $\tan(\pi/2)$ .

Answers

Fully worked answers

Show the answers

(a) $135\pi/180=\boxed{3\pi/4}$ . (b) $300\pi/180=\boxed{5\pi/3}$ . (c) $(3\pi/4)(180/\pi)=\boxed{135^\circ}$ . (d) $(11\pi/6)(180/\pi)=\boxed{330^\circ}$ .
In tenths of $\pi$ , the quadrant boundaries are $0,5\pi/10,10\pi/10,15\pi/10,20\pi/10$ . The angles are respectively $2\pi/10,8\pi/10,12\pi/10,18\pi/10$ . Thus: (a) I, signs $(+,+,+)$ ; (b) II, $(+,−,−)$ ; (c) III, $(−,−,+)$ ; (d) IV, $(−,+,−)$ , in sin/cos/tan order.
(a) Below the horizontal axis: III or IV. (b) Left of the vertical axis: II or III. (c) Sine and cosine have the same sign: I or III. Both sine and cosine negative places the point in III.
(a) Quadrant I; reference $\pi/3$ ; positive sine: $\boxed{\sqrt3/2}$ . (b) Quadrant III; reference $4\pi/3-\pi=\pi/3$ ; negative sine: $\boxed{-\sqrt3/2}$ . (c) Quadrant IV; reference $2\pi-7\pi/4=\pi/4$ ; positive cosine: $\boxed{\sqrt2/2}$ . (d) Quadrant II; reference $\pi-2\pi/3=\pi/3$ ; negative tangent: $\boxed{-\sqrt3}$ .
Quadrant IV; $\alpha=2\pi-5\pi/3=\boxed{\pi/3}$ . Sine is negative, so $\boxed{\sin(5\pi/3)=-\sqrt3/2}$ . Cosine is positive, so $\boxed{\cos(5\pi/3)=1/2}$ .
Subtract one full turn: $3\pi/2-2\pi=\boxed{-\pi/2}$ , which lies in the interval. The symbol $\le$ includes $\boxed{-\pi}$ ; the symbol $<$ excludes $\pi$ , even though these give the same leftward ray.
The first answer may come from dropping $\pi$ and treating the angle as its sine. The second may come from using cosine of the reference angle. The angle is in II, reference $\pi-2\pi/3=\pi/3$ , and sine is positive: $\boxed{\sqrt3/2}$ . These are possible explanations, not diagnoses.
(a) Bottom point $(0,-1)$ : sine is its vertical coordinate, $\boxed{-1}$ . (b) Left point $(-1,0)$ : cosine is its horizontal coordinate, $\boxed{-1}$ . (c) Top point $(0,1)$ : tangent would be $1/0$ , so it is undefined. Next step: explain one answer using the unit circle rather than the sign table alone. Success here prepares you for all trigonometric solutions in an interval; it does not establish mastery of solving all trig equations in an interval.

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