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Precalculus / LEVEL 1 · DIFFICULTY 1/5

Functions and Circular Measurement

Connect domains, radians and coordinates.

3 stages · 24 practice problems · two 6-question assessment forms

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MINI QUEST PC 1.1Real Function DomainsRead the lesson
MINI QUEST PC 1.2Radian MeasureRead the lesson
MINI QUEST PC 1.3Trigonometry from CoordinatesRead the lesson
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STAGE PC 1.1 · 8 PRACTICE PROBLEMS · 2 NEW-CONTEXT APPLICATIONS

Real Function Domains

Goal: Understand and apply real function domains.

Before you begin: Intermediate algebra, coordinate geometry, radicals and function notation.

Understand the idea

A formula can contain several simultaneous restrictions. A square root requires a nonnegative radicand, while a denominator must remain nonzero even at an otherwise permitted input.

\displaystyle \sqrt{g}\, \left(x\right)

requires

\displaystyle g\, \left(x\right) \ge 0

; h(x) in a denominator requires

\displaystyle h\, \left(x\right) \ne 0

Choose and carry out a method

Solve every restriction separately and intersect the permitted sets. A removed denominator zero may create a hole inside an interval of allowed square-root inputs.

Check the reasoning

Test an interior allowed value, the square-root boundary and each excluded point. Simplifying a formula does not automatically restore an input forbidden originally.

WORKED EXAMPLE 1

For $\displaystyle f\, \left(x\right) = \frac{\sqrt{x - 4}}{8 - x}$ , the real domain is [4, $\displaystyle \infty$ ) with one point removed. Which point?

Both the square root and denominator impose restrictions.
$\displaystyle x - 4 \ge 0$ and $\displaystyle 8 - x \ne 0$ .
Exclude $\displaystyle x = 8$ ; it already lies in the square-root range.

WORKED EXAMPLE 2

For $\displaystyle f\, \left(x\right) = \frac{\sqrt{x - 5}}{9 - x}$ , the real domain is [5, $\displaystyle \infty$ ) with one point removed. Which point?

Both the square root and denominator impose restrictions.
$\displaystyle x - 5 \ge 0$ and $\displaystyle 9 - x \ne 0$ .
Exclude $\displaystyle x = 9$ ; it already lies in the square-root range.

Common pitfalls

Possible mix-up: Satisfying the square root is enough.

Every denominator restriction must also hold.

Possible mix-up: A correct numerical answer alone explains the method.

State the governing relationship and check the conditions described above.

Explain it to yourself

Why can a function’s domain be an interval with a single point removed?

Preview the eight practice prompts

Problem 1
For $\displaystyle f\, \left(x\right) = \frac{\sqrt{x - 7}}{11 - x}$ , the real domain is [7, $\displaystyle \infty$ ) with one point removed. Which point?
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Problem 2
For $\displaystyle f\, \left(x\right) = \frac{\sqrt{x - 8}}{12 - x}$ , the real domain is [8, $\displaystyle \infty$ ) with one point removed. Which point?
Practise this exact problem →
Problem 3
For $\displaystyle f\, \left(x\right) = \frac{\sqrt{x - 9}}{13 - x}$ , the real domain is [9, $\displaystyle \infty$ ) with one point removed. Which point?
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Problem 4
For $\displaystyle f\, \left(x\right) = \frac{\sqrt{x - 10}}{14 - x}$ , the real domain is [10, $\displaystyle \infty$ ) with one point removed. Which point?
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Problem 5
For $\displaystyle f\, \left(x\right) = \frac{\sqrt{x - 11}}{15 - x}$ , the real domain is [11, $\displaystyle \infty$ ) with one point removed. Which point?
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Problem 6
For $\displaystyle f\, \left(x\right) = \frac{\sqrt{x - 12}}{16 - x}$ , the real domain is [12, $\displaystyle \infty$ ) with one point removed. Which point?
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Problem 7New context
A real-valued sensor model is $\displaystyle S\, \left(x\right) = \frac{\sqrt{x - 13}}{17 - x}$ . After enforcing $\displaystyle x \ge 13$ , which input must still be excluded?
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Problem 8New context
A real-valued sensor model is $\displaystyle S\, \left(x\right) = \frac{\sqrt{x - 14}}{18 - x}$ . After enforcing $\displaystyle x \ge 14$ , which input must still be excluded?
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Open stage PC 1.1 in the student workspace →

STAGE PC 1.2 · 8 PRACTICE PROBLEMS · 2 NEW-CONTEXT APPLICATIONS

Radian Measure

Useful preparation: Real Function Domains

Goal: Understand and apply radian measure.

Before you begin: Real Function Domains

Understand the idea

An angle measured in radians is the ratio of intercepted arc length to radius. One complete circumference gives $\displaystyle 2\, \pi$ radians, making radians a natural scale for circular motion.

\displaystyle \theta

radians=

\displaystyle \theta

degrees·

\displaystyle \frac{\pi}{180}

Choose and carry out a method

Convert degrees by multiplying by $\displaystyle \frac{\pi}{180}$ . If asked for k in $\displaystyle k\, \pi$ , provide only that exact coefficient.

Check the reasoning

A half-turn is $\displaystyle \pi$ , a quarter-turn is $\displaystyle \frac{\pi}{2}$ and a full turn is $\displaystyle 2\, \pi$ . These landmarks detect reversed conversion factors.

WORKED EXAMPLE 1

Convert $\displaystyle 45^{\circ}$ to radians. Write your answer as $\displaystyle k\, \pi$ and enter k only.

A half revolution is both 180 degrees and $\displaystyle \pi$ radians.
Radians= $\displaystyle \frac{45 \cdot \pi}{180}$ .
The coefficient of $\displaystyle \pi$ is $\displaystyle \frac{1}{4}$ . Angles may exceed one revolution.

$\displaystyle \frac{1}{4}$

WORKED EXAMPLE 2

Convert $\displaystyle 60^{\circ}$ to radians. Write your answer as $\displaystyle k\, \pi$ and enter k only.

A half revolution is both 180 degrees and $\displaystyle \pi$ radians.
Radians= $\displaystyle \frac{60 \cdot \pi}{180}$ .
The coefficient of $\displaystyle \pi$ is $\displaystyle \frac{1}{3}$ . Angles may exceed one revolution.

$\displaystyle \frac{1}{3}$

Common pitfalls

Possible mix-up: Multiply degrees by $\displaystyle \frac{180}{\pi}$ .

That converts radians to degrees; reverse the factor for this direction.

Possible mix-up: A correct numerical answer alone explains the method.

State the governing relationship and check the conditions described above.

Explain it to yourself

Why does the arc-length-to-radius ratio stay unchanged when a circle is enlarged?

Preview the eight practice prompts

Problem 1
Convert $\displaystyle 90^{\circ}$ to radians. Write your answer as $\displaystyle k\, \pi$ and enter k only.
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Problem 2
Convert $\displaystyle 105^{\circ}$ to radians. Write your answer as $\displaystyle k\, \pi$ and enter k only.
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Problem 3
Convert $\displaystyle 120^{\circ}$ to radians. Write your answer as $\displaystyle k\, \pi$ and enter k only.
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Problem 4
Convert $\displaystyle 135^{\circ}$ to radians. Write your answer as $\displaystyle k\, \pi$ and enter k only.
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Problem 5
Convert $\displaystyle 150^{\circ}$ to radians. Write your answer as $\displaystyle k\, \pi$ and enter k only.
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Problem 6
Convert $\displaystyle 165^{\circ}$ to radians. Write your answer as $\displaystyle k\, \pi$ and enter k only.
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Problem 7New context
A wheel rotates through $\displaystyle 180^{\circ}$ . Its rotation is $\displaystyle k\, \pi$ radians. Find k.
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Problem 8New context
A wheel rotates through $\displaystyle 195^{\circ}$ . Its rotation is $\displaystyle k\, \pi$ radians. Find k.
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Open stage PC 1.2 in the student workspace →

STAGE PC 1.3 · 8 PRACTICE PROBLEMS · 2 NEW-CONTEXT APPLICATIONS

Trigonometry from Coordinates

Useful preparation: Radian Measure

Goal: Understand and apply trigonometry from coordinates.

Before you begin: Radian Measure

ONE ROTATION, TWO COORDINATES

From the unit circle to a wave

Parent relation: $\displaystyle x^{2} + y^{2} = 1$ . The radius is 1. Turn the radius and follow the horizontal cosine and vertical sine coordinates.

Trace one complete rotation

$\displaystyle \theta = 0$ rad; $\displaystyle \operatorname{cos}\, \theta = 1$ ; $\displaystyle \operatorname{sin}\, \theta = 0$

The solid purple curve records $\displaystyle \operatorname{sin}\, \theta$ as the angle increases from 0 to $\displaystyle 2\, \pi$ .

The dashed orange curve records $\displaystyle \operatorname{cos}\, \theta$ . Both repeat after one complete turn.

Understand the idea

A point on an angle’s terminal ray determines a right triangle with the axes. Dividing its coordinates by its radius produces cosine and sine, with signs determined by the quadrant.

\displaystyle \operatorname{sin}\, \theta = \frac{y}{r}

;

\displaystyle \operatorname{cos}\, \theta = \frac{x}{r}

Choose and carry out a method

Compute $\displaystyle r = \sqrt{x^{2} + y^{2}}$ . Use $\displaystyle \operatorname{sin}\, \theta = \frac{y}{r}$ and $\displaystyle \operatorname{cos}\, \theta = \frac{x}{r}$ ; retain the coordinate signs rather than replacing them with side-length signs.

Check the reasoning

The normalized coordinates satisfy $\displaystyle \operatorname{sin}^{2}\, \theta + \operatorname{cos}^{2}\, \theta = 1$ . In quadrant II, sine is positive and cosine is negative.

WORKED EXAMPLE 1

The terminal ray of $\displaystyle \theta$ passes through (-8,6) in quadrant II. Find $\displaystyle \operatorname{sin}\, \theta$ as a fraction.

Sine is y divided by distance from the origin, with the sign of y.
$\displaystyle r^{2} = 8^{2} + 6^{2} = 100$ , so $\displaystyle r = 10$ .
$\displaystyle \operatorname{sin}\, \theta = \frac{6}{10} = \frac{3}{5}$ , positive in quadrant II.

$\displaystyle \frac{3}{5}$

WORKED EXAMPLE 2

The terminal ray of $\displaystyle \theta$ passes through (-15,8) in quadrant II. Find $\displaystyle \operatorname{sin}\, \theta$ as a fraction.

Sine is y divided by distance from the origin, with the sign of y.
$\displaystyle r^{2} = 15^{2} + 8^{2} = 289$ , so $\displaystyle r = 17$ .
$\displaystyle \operatorname{sin}\, \theta = \frac{8}{17} = \frac{8}{17}$ , positive in quadrant II.

$\displaystyle \frac{8}{17}$

Common pitfalls

Possible mix-up: Both trigonometric values must be positive because lengths are positive.

Signed coordinates determine quadrant signs.

Possible mix-up: A correct numerical answer alone explains the method.

State the governing relationship and check the conditions described above.

Explain it to yourself

How does reflecting a point across the x-axis affect its sine and cosine?

Preview the eight practice prompts

Problem 1
The terminal ray of $\displaystyle \theta$ passes through (-35,12) in quadrant II. Find $\displaystyle \operatorname{sin}\, \theta$ as a fraction.
Practise this exact problem →
Problem 2
The terminal ray of $\displaystyle \theta$ passes through (-48,14) in quadrant II. Find $\displaystyle \operatorname{sin}\, \theta$ as a fraction.
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Problem 3
The terminal ray of $\displaystyle \theta$ passes through (-63,16) in quadrant II. Find $\displaystyle \operatorname{sin}\, \theta$ as a fraction.
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Problem 4
The terminal ray of $\displaystyle \theta$ passes through (-80,18) in quadrant II. Find $\displaystyle \operatorname{sin}\, \theta$ as a fraction.
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Problem 5
The terminal ray of $\displaystyle \theta$ passes through (-99,20) in quadrant II. Find $\displaystyle \operatorname{sin}\, \theta$ as a fraction.
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Problem 6
The terminal ray of $\displaystyle \theta$ passes through (-120,22) in quadrant II. Find $\displaystyle \operatorname{sin}\, \theta$ as a fraction.
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Problem 7New context
A rotating arm points from the origin toward (-143,24). What is the vertical component of its unit direction vector?
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Problem 8New context
A rotating arm points from the origin toward (-168,26). What is the vertical component of its unit direction vector?
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Open stage PC 1.3 in the student workspace →

Browse another level →

Functions and Circular Measurement

Real Function Domains

Understand the idea

Choose and carry out a method

Check the reasoning

For f (x)=x−48−x\displaystyle f\, \left(x\right) = \frac{\sqrt{x - 4}}{8 - x}f(x)=8−xx−4​​, the real domain is [4,∞\displaystyle \infty∞) with one point removed. Which point?

For f (x)=x−59−x\displaystyle f\, \left(x\right) = \frac{\sqrt{x - 5}}{9 - x}f(x)=9−xx−5​​, the real domain is [5,∞\displaystyle \infty∞) with one point removed. Which point?

Common pitfalls

Explain it to yourself

Radian Measure

Understand the idea

Choose and carry out a method

Check the reasoning

Convert 45∘\displaystyle 45^{\circ}45∘ to radians. Write your answer as k π\displaystyle k\, \pikπ and enter k only.

Convert 60∘\displaystyle 60^{\circ}60∘ to radians. Write your answer as k π\displaystyle k\, \pikπ and enter k only.

Common pitfalls

Explain it to yourself

Trigonometry from Coordinates

From the unit circle to a wave

Understand the idea

Choose and carry out a method

Check the reasoning

The terminal ray of θ\displaystyle \thetaθ passes through (-8,6) in quadrant II. Find sin⁡ θ\displaystyle \operatorname{sin}\, \thetasinθ as a fraction.

The terminal ray of θ\displaystyle \thetaθ passes through (-15,8) in quadrant II. Find sin⁡ θ\displaystyle \operatorname{sin}\, \thetasinθ as a fraction.

Common pitfalls

Explain it to yourself

For $\displaystyle f\, \left(x\right) = \frac{\sqrt{x - 4}}{8 - x}$ , the real domain is [4, $\displaystyle \infty$ ) with one point removed. Which point?

For $\displaystyle f\, \left(x\right) = \frac{\sqrt{x - 5}}{9 - x}$ , the real domain is [5, $\displaystyle \infty$ ) with one point removed. Which point?

Convert $\displaystyle 45^{\circ}$ to radians. Write your answer as $\displaystyle k\, \pi$ and enter k only.

Convert $\displaystyle 60^{\circ}$ to radians. Write your answer as $\displaystyle k\, \pi$ and enter k only.

The terminal ray of $\displaystyle \theta$ passes through (-8,6) in quadrant II. Find $\displaystyle \operatorname{sin}\, \theta$ as a fraction.

The terminal ray of $\displaystyle \theta$ passes through (-15,8) in quadrant II. Find $\displaystyle \operatorname{sin}\, \theta$ as a fraction.