Precalculus · F.TF.A.3

Exact values on the unit circle

A lighthouse beam turns until it points along \displaystyle \frac{4\pi}{3}, or straight down the page at \displaystyle \frac{3\pi}{2}. At which angles can all six trigonometric values be written exactly, with no calculator, and how are they read off the circle?

You need

Reference anglesPrecalculus Secant, cosecant and cotangent on the unit circlePrecalculus Trig values of the special right trianglesGeometry

Unlocks

Law of cosinesPrecalculus Solving basic trig equationsPrecalculus Sine and tangent of a sum or differencePrecalculus

Where you are 8 in The unit circle

Trigonometry7

The unit circle8

▴ 3 above 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

Exact values on the unit circle

When you reach for it

On the unit circle every angle \theta has a terminal point (x, y), with \cos\theta = x and \sin\theta = y (Trig functions on the unit circle), and the other four functions are built from the same two numbers (Secant, cosecant and cotangent on the unit circle). For most angles those coordinates are decimals a calculator has to supply. For two families of angles they are exact numbers you can write down by hand:

the quadrantal angles, the multiples of \displaystyle \frac{\pi}{2} (or 90^\circ), whose terminal side lies on an axis;
the multiples of \displaystyle \frac{\pi}{6} and of \displaystyle \frac{\pi}{4} (or 30^\circ and 45^\circ), whose reference angle is \displaystyle \frac{\pi}{6}, \displaystyle \frac{\pi}{4} or \displaystyle \frac{\pi}{3}.

You reach for this whenever a trig value is wanted at one of these angles: to evaluate an expression exactly, to place a point a given distance away in a given direction, to check a calculator, or as the first step in solving an equation such as \displaystyle \cos\theta = -\frac{1}{2}. Exact values also carry structure a decimal hides: \displaystyle \tan\frac{\pi}{3} = \sqrt{3} says the line at that angle has slope \displaystyle \sqrt{3}, which 1.732 only approximates.

The lesson runs over two days. Part 1 places the sixteen exact points on the circle. Part 2 reads all six functions at any of those angles.

The procedure

Part 1: the quadrantal angles

A quadrantal angle ends on an axis, so its terminal point is where that axis meets the circle of radius 1. Going counterclockwise from the positive x-axis the four points are

\theta	point	\cos\theta	\sin\theta
0 = 0^\circ	(1, 0)	1	0
\displaystyle \frac{\pi}{2} = 90^\circ	(0, 1)	0	1
\pi = 180^\circ	(-1, 0)	-1	0
\displaystyle \frac{3\pi}{2} = 270^\circ	(0, -1)	0	-1

No triangle is involved: one coordinate is 0 because the point is on an axis, and the other is \pm 1 because the point is one unit from the origin. Any other quadrantal angle is one of these four plus or minus whole turns of 2\pi, so it shares their values. To place it, subtract (or add) whole turns until the angle lies in 0 \le \theta < 2\pi, as in Reference angles.

Part 1: the three angles of Quadrant I

The terminal point of an acute angle \theta is the top corner of a right triangle whose hypotenuse is a radius, of length 1, and whose legs run along and up from the x-axis. Since the hypotenuse is 1, the horizontal leg is \cos\theta and the vertical leg is \sin\theta, exactly the values of the special right triangles:

\theta	terminal point
\displaystyle \frac{\pi}{6} = 30^\circ	\displaystyle \left(\frac{\sqrt{3}}{2}, \frac{1}{2}\right)
\displaystyle \frac{\pi}{4} = 45^\circ	\displaystyle \left(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right)
\displaystyle \frac{\pi}{3} = 60^\circ	\displaystyle \left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right)

Figure 1The three Quadrant I points. A ray near the x-axis has the larger x-coordinate; a ray near the y-axis has the larger y-coordinate.

Two checks keep the table straight. Every point is on the unit circle, so its coordinates satisfy x^2 + y^2 = 1: for instance \displaystyle \frac{3}{4} + \frac{1}{4} = 1. And the coordinate that is larger in size belongs to the axis the ray is nearer: at \displaystyle \frac{\pi}{6} the ray is close to the x-axis, so \displaystyle x = \frac{\sqrt{3}}{2} is the large one; at \displaystyle \frac{\pi}{3} it is close to the y-axis, so \displaystyle y = \frac{\sqrt{3}}{2} is. At \displaystyle \frac{\pi}{4}, halfway, the two are equal.

Part 1: the other twelve points

Every multiple of \displaystyle \frac{\pi}{6} or \displaystyle \frac{\pi}{4} that is not quadrantal has reference angle \displaystyle \frac{\pi}{6}, \displaystyle \frac{\pi}{4} or \displaystyle \frac{\pi}{3}. Its terminal point is the matching Quadrant I point reflected into its own quadrant: the same two sizes, in the same order, with signs set by the quadrant (left of the y-axis x < 0, below the x-axis y < 0, as in Signs of the trig functions by quadrant). With the four axis points that gives sixteen points in one turn.

Figure 2The sixteen angles in one turn whose terminal points have exact coordinates built from 0, \pm 1, \displaystyle \pm\frac{1}{2}, \displaystyle \pm\frac{\sqrt{2}}{2} and \displaystyle \pm\frac{\sqrt{3}}{2}.

Which reference angle an angle has can be read from its fraction once it is inside one turn. An angle \displaystyle \frac{k\pi}{4} with k odd has reference angle \displaystyle \frac{\pi}{4}, so both coordinates have size \displaystyle \frac{\sqrt{2}}{2}. An angle \displaystyle \frac{k\pi}{6} in lowest terms (so k odd, not a multiple of 3) has reference angle \displaystyle \frac{\pi}{6}. An angle \displaystyle \frac{k\pi}{3} in lowest terms has reference angle \displaystyle \frac{\pi}{3}. Reduce the fraction first: \displaystyle \frac{8\pi}{6} is \displaystyle \frac{4\pi}{3}, a \displaystyle \frac{\pi}{3} angle.

Part 2: all six functions

With the terminal point (x, y) in hand, the other four functions are quotients:

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

At a quadrantal angle one coordinate is 0. Dividing by it is impossible, so two of the four are undefined and the other two are 0 or \pm 1:

	0	\displaystyle \frac{\pi}{2}	\pi	\displaystyle \frac{3\pi}{2}
\sin	0	1	0	-1
\cos	1	0	-1	0
\tan	0	none	0	none
\cot	none	0	none	0
\sec	1	none	-1	none
\csc	none	1	none	-1

In the table, “none” marks a value that is undefined: the function has no value at that angle.

A 0 on top of a fraction gives 0; a 0 underneath gives no value at all.

At any other angle of the two families both coordinates are nonzero and all six exist. Their sizes depend only on the reference angle \theta'. Dividing the sizes from the Quadrant I table, and clearing roots from denominators as in Rationalizing a denominator (for instance \displaystyle \frac{1}{\sqrt{3}/2} = \frac{2}{\sqrt{3}} = \frac{2\sqrt{3}}{3}):

\theta'	\displaystyle \frac{\pi}{6}	\displaystyle \frac{\pi}{4}	\displaystyle \frac{\pi}{3}
\sin	\displaystyle \frac{1}{2}	\displaystyle \frac{\sqrt{2}}{2}	\displaystyle \frac{\sqrt{3}}{2}
\cos	\displaystyle \frac{\sqrt{3}}{2}	\displaystyle \frac{\sqrt{2}}{2}	\displaystyle \frac{1}{2}
\tan	\displaystyle \frac{\sqrt{3}}{3}	1	\displaystyle \sqrt{3}
\cot	\displaystyle \sqrt{3}	1	\displaystyle \frac{\sqrt{3}}{3}
\sec	\displaystyle \frac{2\sqrt{3}}{3}	\displaystyle \sqrt{2}	2
\csc	2	\displaystyle \sqrt{2}	\displaystyle \frac{2\sqrt{3}}{3}

The signs come from the quadrant, and each reciprocal keeps the sign of its partner: secant goes with cosine, cosecant with sine, and cotangent with tangent. Tangent and cotangent are positive exactly where x and y agree in sign, Quadrants I and III.

Worked examples

Example 1

Part 1: Riya’s beam at two quadrantal angles

Riya’s lighthouse lamp sits at the origin with its beam pointing due east, along the positive x-axis, with north along the positive y-axis. It turns \displaystyle \frac{3\pi}{2} counterclockwise. In which direction does the beam point, and what are \displaystyle \cos\frac{3\pi}{2} and \displaystyle \sin\frac{3\pi}{2}? Answer the same questions if it had instead turned 5\pi clockwise.

Solution

Three quarter turns counterclockwise from east pass north and west and stop at south, on the negative y-axis. The terminal point is where that half-axis meets the unit circle, (0, -1), so

\cos\frac{3\pi}{2} = 0, \qquad \sin\frac{3\pi}{2} = -1

A clockwise turn is a negative angle, -5\pi. Adding three full turns (6\pi) moves the terminal side nowhere and gives

-5\pi + 6\pi = \pi

so the beam ends pointing due west, at (-1, 0). Then \cos(-5\pi) = -1 and \sin(-5\pi) = 0. Every odd multiple of \pi ends there; every even multiple ends due east.

Example 2

Part 1: two terminal points far from one turn

Find the terminal points of \displaystyle -\frac{19\pi}{6} and of \displaystyle \frac{21\pi}{4}.

Solution

For \displaystyle -\frac{19\pi}{6}, a full turn is \displaystyle \frac{12\pi}{6}. One turn added gives \displaystyle -\frac{7\pi}{6}, still negative, so add two:

-\frac{19\pi}{6} + \frac{24\pi}{6} = \frac{5\pi}{6}

The fraction \displaystyle \frac{5\pi}{6} is in lowest terms with denominator 6, so the reference angle is \displaystyle \frac{\pi}{6}: the ray is near the x-axis and \displaystyle |x| = \frac{\sqrt{3}}{2}, \displaystyle |y| = \frac{1}{2}. Since \displaystyle \frac{5\pi}{6} is between \displaystyle \frac{\pi}{2} and \pi, the point is in Quadrant II, left of the y-axis and above the x-axis. The terminal point is \displaystyle \left(-\frac{\sqrt{3}}{2}, \frac{1}{2}\right).

For \displaystyle \frac{21\pi}{4}, two full turns are \displaystyle \frac{16\pi}{4}, and

\frac{21\pi}{4} - \frac{16\pi}{4} = \frac{5\pi}{4}

an odd number of quarters of \pi, so both coordinates have size \displaystyle \frac{\sqrt{2}}{2}. The angle is between \pi and \displaystyle \frac{3\pi}{2}, Quadrant III, where both are negative. The terminal point is \displaystyle \left(-\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2}\right).

Example 3

Part 1: Shreya’s boat

Shreya keeps a lighthouse whose beam starts due east and turns counterclockwise at a steady \displaystyle \frac{\pi}{6} radians per second. A boat sits 4 miles from the lighthouse, exactly where the beam points after 22 seconds. With the lighthouse at the origin, east along the positive x-axis and north along the positive y-axis, find the boat’s exact position.

Solution

In 22 seconds the beam turns

22 \cdot \frac{\pi}{6} = \frac{22\pi}{6} = \frac{11\pi}{3}

Subtracting one turn, \displaystyle \frac{11\pi}{3} - \frac{6\pi}{3} = \frac{5\pi}{3}, between \displaystyle \frac{3\pi}{2} and 2\pi, so Quadrant IV. In lowest terms the denominator is 3, so the reference angle is \displaystyle \frac{\pi}{3} and the terminal point is \displaystyle \left(\frac{1}{2}, -\frac{\sqrt{3}}{2}\right): right of the y-axis, below the x-axis.

The boat is on the circle of radius 4, and scaling the unit circle by 4 scales every point by 4:

4\left(\frac{1}{2}, -\frac{\sqrt{3}}{2}\right) = \left(2, -2\sqrt{3}\right)

The boat is 2 miles east and \displaystyle 2\sqrt{3} \approx 3.46 miles south of the lighthouse. Check: its distance is \displaystyle \sqrt{4 + 12} = 4 miles, as it must be.

Example 4

Part 2: all six values at 4 pi over 3

Cuauhtemoc’s lighthouse beam stops at \displaystyle \frac{4\pi}{3}. Find all six trigonometric values of \displaystyle \frac{4\pi}{3}.

Solution

The angle is already inside one turn, and \displaystyle \pi < \frac{4\pi}{3} < \frac{3\pi}{2}, so it is in Quadrant III. It passes the negative x-axis by \displaystyle \frac{4\pi}{3} - \pi = \frac{\pi}{3}, its reference angle.

Figure 3The beam at \displaystyle \frac{4\pi}{3} makes a reference angle of \displaystyle \frac{\pi}{3} with the negative x-axis, so its terminal point is the \displaystyle \frac{\pi}{3} point reflected into Quadrant III.

The \displaystyle \frac{\pi}{3} point is \displaystyle \left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right), and in Quadrant III both coordinates are negative, so the terminal point is \displaystyle \left(-\frac{1}{2}, -\frac{\sqrt{3}}{2}\right). Reading and dividing:

\begin{aligned} \sin\frac{4\pi}{3} &= -\frac{\sqrt{3}}{2} \\ \cos\frac{4\pi}{3} &= -\frac{1}{2} \\ \tan\frac{4\pi}{3} &= \frac{-\sqrt{3}/2}{-1/2} = \sqrt{3} \\ \cot\frac{4\pi}{3} &= \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3} \\ \sec\frac{4\pi}{3} &= \frac{1}{-1/2} = -2 \\ \csc\frac{4\pi}{3} &= -\frac{2}{\sqrt{3}} = -\frac{2\sqrt{3}}{3} \end{aligned}

Tangent and cotangent are positive, as they must be in Quadrant III, and the sizes match the \displaystyle \frac{\pi}{3} row of the table.

Example 5

Part 2: three values in degrees

Find \sec 495^\circ, \csc(-270^\circ) and \cot 300^\circ exactly.

Solution

495^\circ - 360^\circ = 135^\circ, in Quadrant II with reference angle 45^\circ. The cosine there is \displaystyle -\frac{\sqrt{2}}{2}, so

\sec 495^\circ = \frac{1}{-\sqrt{2}/2} = -\frac{2}{\sqrt{2}} = -\sqrt{2}

-270^\circ + 360^\circ = 90^\circ, a quadrantal angle with terminal point (0, 1). The cosecant divides by y = 1, so it exists: \displaystyle \csc(-270^\circ) = \frac{1}{1} = 1. (Its tangent and secant, which divide by x = 0, do not.)

300^\circ is in Quadrant IV with reference angle 60^\circ, terminal point \displaystyle \left(\frac{1}{2}, -\frac{\sqrt{3}}{2}\right). Then

\cot 300^\circ = \frac{1/2}{-\sqrt{3}/2} = -\frac{1}{\sqrt{3}} = -\frac{\sqrt{3}}{3}

Example 6

Part 2: an expression with a quadrantal term

Evaluate exactly:

2\sec\frac{5\pi}{3}\cot\frac{3\pi}{4} - \csc^2\frac{7\pi}{6} + \cot\frac{3\pi}{2}

Solution

Take the four values one at a time. \displaystyle \frac{5\pi}{3} is the Quadrant IV point \displaystyle \left(\frac{1}{2}, -\frac{\sqrt{3}}{2}\right), so \displaystyle \sec\frac{5\pi}{3} = \frac{1}{1/2} = 2. \displaystyle \frac{3\pi}{4} is the Quadrant II point \displaystyle \left(-\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right), equal sizes with opposite signs, so \displaystyle \cot\frac{3\pi}{4} = -1. \displaystyle \frac{7\pi}{6} is the Quadrant III point \displaystyle \left(-\frac{\sqrt{3}}{2}, -\frac{1}{2}\right), so \displaystyle \csc\frac{7\pi}{6} = -2, and \csc^2 means the square of the value, (-2)^2 = 4. Finally \displaystyle \frac{3\pi}{2} ends at (0, -1), and \displaystyle \cot\frac{3\pi}{2} = \frac{0}{-1} = 0: the zero is on top, so the value exists. Putting them together,

2(2)(-1) - 4 + 0 = -8

Example 7

Part 2: Cuauhtemoc’s breakwater

Cuauhtemoc’s lighthouse stands at the origin, 3 miles due south of a long straight breakwater that runs east and west along the line y = 3. Where does the beam strike the breakwater when it points along \displaystyle \frac{2\pi}{3}, and how far is that spot from the lamp? What happens at \theta = \pi?

Solution

The beam leaves along the terminal side of \theta, through the terminal point (x, y). Multiplying that point by \displaystyle \frac{3}{y} moves it along the same ray (for y > 0) to height 3:

\frac{3}{y}(x, y) = \left(3\cdot\frac{x}{y},\ 3\right) = (3\cot\theta,\ 3)

and its distance from the origin is \displaystyle \frac{3}{y} times 1, that is 3\csc\theta.

At \displaystyle \frac{2\pi}{3} (Quadrant II, reference angle \displaystyle \frac{\pi}{3}) the terminal point is \displaystyle \left(-\frac{1}{2}, \frac{\sqrt{3}}{2}\right), so

\begin{aligned} \cot\frac{2\pi}{3} &= \frac{-1/2}{\sqrt{3}/2} = -\frac{\sqrt{3}}{3} \\ \csc\frac{2\pi}{3} &= \frac{2}{\sqrt{3}} = \frac{2\sqrt{3}}{3} \end{aligned}

The beam strikes the breakwater at \displaystyle \left(-\sqrt{3}, 3\right), about 1.73 miles west of the point due north of the lamp, at a distance of \displaystyle 3 \cdot \frac{2\sqrt{3}}{3} = 2\sqrt{3} \approx 3.46 miles. Check: \displaystyle \sqrt{3 + 9} = \sqrt{12} = 2\sqrt{3}.

At \theta = \pi the beam points due west along the x-axis, parallel to the breakwater, and never reaches it. The terminal point (-1, 0) has y = 0, so \cot\pi and \csc\pi are undefined, which is the same fact in symbols.

Boundaries

Only two families of angles. These exact values belong to the multiples of \displaystyle \frac{\pi}{6} and \displaystyle \frac{\pi}{4}. Most other angles, such as \displaystyle \frac{\pi}{5} or 1 radian, need a calculator here; some, such as \displaystyle \frac{\pi}{12}, can be found exactly from the sum and difference formulas, but not from this table.

Quadrantal angles have no reference angle. Their values come straight from the four axis points, and at each one two of the six functions are undefined: tangent and secant where x = 0, cotangent and cosecant where y = 0.

Undefined is not zero. \displaystyle \cot\frac{\pi}{2} = 0, while \displaystyle \tan\frac{\pi}{2} has no value.

Radians and degrees. \displaystyle \frac{\pi}{6} and 30^\circ are the same angle, but \sin 30, read in radians, is not \displaystyle \frac{1}{2}.

Where this goes next

Exact values are the raw material for the rest of trigonometry. Run backward, they solve equations such as \sec\theta = -2 (Solving basic trig equations). Combined in the sum and difference formulas they give exact values at new angles, such as \displaystyle \frac{\pi}{12} = \frac{\pi}{3} - \frac{\pi}{4}. Placing a point at distance b in direction C as (b\cos C, b\sin C), as with Shreya’s boat, is the step that turns the distance formula into the law of cosines. And the sixteen points are the landmarks used to sketch the graphs of all six functions, one period at a time (periodicity).

Exercises

Part 1: quadrantal angles

Find each value exactly.

\cos(-7\pi)
\displaystyle \sin\left(-\frac{5\pi}{2}\right)
\cos(-810^\circ)
\displaystyle \sin\frac{9\pi}{2}

Part 1: terminal points

Find the terminal point on the unit circle.

\displaystyle \frac{31\pi}{6}
\displaystyle -\frac{7\pi}{4}
\displaystyle \frac{14\pi}{3}
510^\circ
-135^\circ

Part 1: using the points

Find every \theta with 0 \le \theta < 2\pi, among the multiples of \displaystyle \frac{\pi}{6} and \displaystyle \frac{\pi}{4}, with \displaystyle \sin\theta = -\frac{1}{2}.
Riya’s beam starts due east and turns \displaystyle \frac{3\pi}{4} clockwise. A boat sits 5 miles out along it. With the lamp at the origin, east along the positive x-axis and north along the positive y-axis, find the boat’s exact position and round it to the nearest hundredth of a mile.
A lighthouse beam starts due east and turns counterclockwise \displaystyle \frac{\pi}{4} every 5 seconds. Where does it point after 50 seconds? Give the cosine and sine of the angle it has turned.

Part 2: all six values

Find all six trigonometric values, writing “undefined” where a value does not exist.

\displaystyle \frac{19\pi}{4}
\displaystyle -\frac{\pi}{6}
-6\pi
-210^\circ

Part 2: single values

\displaystyle \sec\left(-\frac{7\pi}{3}\right)
\displaystyle \csc\frac{13\pi}{4}
\displaystyle \cot\left(-\frac{7\pi}{6}\right)
\displaystyle \tan\left(-\frac{9\pi}{2}\right)

Part 2: expressions and the breakwater

Evaluate exactly: \displaystyle \csc^2\frac{\pi}{4} - \tan\frac{5\pi}{3}\cot\frac{\pi}{6} + \sec\pi.
A lighthouse stands 2 miles due south of a straight breakwater along y = 2. Where does the beam strike the breakwater when it points along \displaystyle \frac{3\pi}{4}, and how far is that spot from the lamp? Give the distance exactly and to the nearest hundredth of a mile.

Reasoning

Explain why, at every multiple of \displaystyle \frac{\pi}{4} that is not a multiple of \displaystyle \frac{\pi}{2}, the tangent and cotangent are \pm 1 and the secant and cosecant are \displaystyle \pm\sqrt{2}.
Shreya writes \displaystyle \csc\frac{5\pi}{6} = \frac{2\sqrt{3}}{3}, because the reference angle is \displaystyle \frac{\pi}{6} and \displaystyle \cos\frac{\pi}{6} = \frac{\sqrt{3}}{2}. Find her mistake and give the correct value.
Explain why, among the angles in 0 \le \theta < 2\pi, \sin\theta = \cos\theta holds only at \displaystyle \frac{\pi}{4} and \displaystyle \frac{5\pi}{4}.

Answers

-1. -7\pi + 8\pi = \pi, terminal point (-1, 0).
-1. \displaystyle -\frac{5\pi}{2} + 4\pi = \frac{3\pi}{2}, terminal point (0, -1).
0. -810^\circ + 1{,}080^\circ = 270^\circ, terminal point (0, -1).
1. \displaystyle \frac{9\pi}{2} - 4\pi = \frac{\pi}{2}, terminal point (0, 1).
\displaystyle \left(-\frac{\sqrt{3}}{2}, -\frac{1}{2}\right). \displaystyle \frac{31\pi}{6} - 4\pi = \frac{7\pi}{6}, Quadrant III with reference angle \displaystyle \frac{\pi}{6}.
\displaystyle \left(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right). \displaystyle -\frac{7\pi}{4} + 2\pi = \frac{\pi}{4}.
\displaystyle \left(-\frac{1}{2}, \frac{\sqrt{3}}{2}\right). \displaystyle \frac{14\pi}{3} - 4\pi = \frac{2\pi}{3}, Quadrant II with reference angle \displaystyle \frac{\pi}{3}.
\displaystyle \left(-\frac{\sqrt{3}}{2}, \frac{1}{2}\right). 510^\circ - 360^\circ = 150^\circ, Quadrant II with reference angle 30^\circ.
\displaystyle \left(-\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2}\right). -135^\circ + 360^\circ = 225^\circ, Quadrant III.
\displaystyle \frac{7\pi}{6} and \displaystyle \frac{11\pi}{6}. A sine of size \displaystyle \frac{1}{2} means reference angle \displaystyle \frac{\pi}{6}; a negative sine means below the x-axis, Quadrants III and IV: \displaystyle \pi + \frac{\pi}{6} and \displaystyle 2\pi - \frac{\pi}{6}.
\displaystyle \left(-\frac{5\sqrt{2}}{2}, -\frac{5\sqrt{2}}{2}\right) \approx (-3.54, -3.54). The angle is \displaystyle -\frac{3\pi}{4}, coterminal with \displaystyle \frac{5\pi}{4}, terminal point \displaystyle \left(-\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2}\right), scaled by 5.
Due north; cosine 0 and sine 1. In 50 seconds it makes 10 turns of \displaystyle \frac{\pi}{4}, so \displaystyle \frac{10\pi}{4} = \frac{5\pi}{2}, and \displaystyle \frac{5\pi}{2} - 2\pi = \frac{\pi}{2}, terminal point (0, 1).
\displaystyle \sin = \frac{\sqrt{2}}{2}, \displaystyle \cos = -\frac{\sqrt{2}}{2}, \tan = -1, \cot = -1, \displaystyle \sec = -\sqrt{2}, \displaystyle \csc = \sqrt{2}. \displaystyle \frac{19\pi}{4} - 4\pi = \frac{3\pi}{4}, Quadrant II.
\displaystyle \sin = -\frac{1}{2}, \displaystyle \cos = \frac{\sqrt{3}}{2}, \displaystyle \tan = -\frac{\sqrt{3}}{3}, \displaystyle \cot = -\sqrt{3}, \displaystyle \sec = \frac{2\sqrt{3}}{3}, \csc = -2. \displaystyle -\frac{\pi}{6} + 2\pi = \frac{11\pi}{6}, Quadrant IV with reference angle \displaystyle \frac{\pi}{6}.
\sin = 0, \cos = 1, \tan = 0, \sec = 1; \cot and \csc undefined. -6\pi is three whole turns, terminal point (1, 0), and y = 0.
\displaystyle \sin = \frac{1}{2}, \displaystyle \cos = -\frac{\sqrt{3}}{2}, \displaystyle \tan = -\frac{\sqrt{3}}{3}, \displaystyle \cot = -\sqrt{3}, \displaystyle \sec = -\frac{2\sqrt{3}}{3}, \csc = 2. -210^\circ + 360^\circ = 150^\circ, Quadrant II.
2. \displaystyle -\frac{7\pi}{3} + 4\pi = \frac{5\pi}{3}, where \displaystyle x = \frac{1}{2}.
\displaystyle -\sqrt{2}. \displaystyle \frac{13\pi}{4} - 2\pi = \frac{5\pi}{4}, where \displaystyle y = -\frac{\sqrt{2}}{2}, and \displaystyle \frac{1}{-\sqrt{2}/2} = -\sqrt{2}.
\displaystyle -\sqrt{3}. \displaystyle -\frac{7\pi}{6} + 2\pi = \frac{5\pi}{6}, terminal point \displaystyle \left(-\frac{\sqrt{3}}{2}, \frac{1}{2}\right), and \displaystyle \frac{x}{y} = -\sqrt{3}.
Undefined. \displaystyle -\frac{9\pi}{2} + 6\pi = \frac{3\pi}{2}, terminal point (0, -1), and the tangent divides by x = 0.
4. \displaystyle \csc^2\frac{\pi}{4} = (\sqrt{2})^2 = 2; \displaystyle \tan\frac{5\pi}{3} = -\sqrt{3} and \displaystyle \cot\frac{\pi}{6} = \sqrt{3}, product -3; \sec\pi = -1. So 2 - (-3) + (-1) = 4.
At (-2, 2), \displaystyle 2\sqrt{2} \approx 2.83 miles from the lamp. The spot is (2\cot\theta, 2) at distance 2\csc\theta; at \displaystyle \frac{3\pi}{4}, \cot = -1 and \displaystyle \csc = \sqrt{2}.
Such an angle has reference angle \displaystyle \frac{\pi}{4}, so \displaystyle |x| = |y| = \frac{\sqrt{2}}{2}. Then \displaystyle \left|\frac{y}{x}\right| = \left|\frac{x}{y}\right| = 1, and \displaystyle \frac{1}{|x|} = \frac{1}{|y|} = \frac{2}{\sqrt{2}} = \sqrt{2}. The signs depend on the quadrant, so each value is \pm 1 or \displaystyle \pm\sqrt{2}.
The cosecant is the reciprocal of the sine, not of the cosine. \displaystyle \sin\frac{5\pi}{6} = \frac{1}{2} (reference angle \displaystyle \frac{\pi}{6}, Quadrant II, above the x-axis), so \displaystyle \csc\frac{5\pi}{6} = 2.
\sin\theta = \cos\theta means the terminal point has y = x, so it lies on the line y = x. That line crosses the unit circle at exactly two points, where 2x^2 = 1: \displaystyle \left(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right) and \displaystyle \left(-\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2}\right), the terminal points of \displaystyle \frac{\pi}{4} and \displaystyle \frac{5\pi}{4}.