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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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  1. MINI QUEST PC 1.1Real Function DomainsRead the lesson
  2. MINI QUEST PC 1.2Radian MeasureRead the lesson
  3. 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.

√g(x) requires g(x)≥0; h(x) in a denominator requires h(x)≠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 f(x)=√(x-4)/(8-x), the real domain is [4,∞) with one point removed. Which point?

  1. Both the square root and denominator impose restrictions.
  2. x-4≥0 and 8-x≠0.
  3. Exclude x=8; it already lies in the square-root range.

8

WORKED EXAMPLE 2

For f(x)=√(x-5)/(9-x), the real domain is [5,∞) with one point removed. Which point?

  1. Both the square root and denominator impose restrictions.
  2. x-5≥0 and 9-x≠0.
  3. Exclude x=9; it already lies in the square-root range.

9

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
  1. For f(x)=√(x-7)/(11-x), the real domain is [7,∞) with one point removed. Which point?
  2. For f(x)=√(x-8)/(12-x), the real domain is [8,∞) with one point removed. Which point?
  3. For f(x)=√(x-9)/(13-x), the real domain is [9,∞) with one point removed. Which point?
  4. For f(x)=√(x-10)/(14-x), the real domain is [10,∞) with one point removed. Which point?
  5. For f(x)=√(x-11)/(15-x), the real domain is [11,∞) with one point removed. Which point?
  6. For f(x)=√(x-12)/(16-x), the real domain is [12,∞) with one point removed. Which point?
  7. A real-valued sensor model is S(x)=√(x-13)/(17-x). After enforcing x≥13, which input must still be excluded? New context
  8. A real-valued sensor model is S(x)=√(x-14)/(18-x). After enforcing x≥14, which input must still be excluded? New context
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 2π radians, making radians a natural scale for circular motion.

θ radians=θ degrees·π/180

Choose and carry out a method

Convert degrees by multiplying by π/180. If asked for k in kπ, provide only that exact coefficient.

Check the reasoning

A half-turn is π, a quarter-turn is π/2 and a full turn is 2π. These landmarks detect reversed conversion factors.

WORKED EXAMPLE 1

Convert 45° to radians. Write your answer as kπ and enter k only.

  1. A half revolution is both 180 degrees and π radians.
  2. Radians=45·π/180.
  3. The coefficient of π is 1/4. Angles may exceed one revolution.

1/4

WORKED EXAMPLE 2

Convert 60° to radians. Write your answer as kπ and enter k only.

  1. A half revolution is both 180 degrees and π radians.
  2. Radians=60·π/180.
  3. The coefficient of π is 1/3. Angles may exceed one revolution.

1/3

Common pitfalls

Possible mix-up: Multiply degrees by 180/π.

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
  1. Convert 90° to radians. Write your answer as kπ and enter k only.
  2. Convert 105° to radians. Write your answer as kπ and enter k only.
  3. Convert 120° to radians. Write your answer as kπ and enter k only.
  4. Convert 135° to radians. Write your answer as kπ and enter k only.
  5. Convert 150° to radians. Write your answer as kπ and enter k only.
  6. Convert 165° to radians. Write your answer as kπ and enter k only.
  7. A wheel rotates through 180°. Its rotation is kπ radians. Find k. New context
  8. A wheel rotates through 195°. Its rotation is kπ radians. Find k. New context
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

(cos θ, sin θ)θxy0
On a unit circle, the point is (cos θ, sin θ). In quadrant II, cosine is negative and sine is positive.

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.

sin θ=y/r; cos θ=x/r

Choose and carry out a method

Compute r=√(x²+y²). Use sin θ=y/r and cos θ=x/r; retain the coordinate signs rather than replacing them with side-length signs.

Check the reasoning

The normalized coordinates satisfy sin²θ+cos²θ=1. In quadrant II, sine is positive and cosine is negative.

WORKED EXAMPLE 1

The terminal ray of θ passes through (-8,6) in quadrant II. Find sin θ as a fraction.

  1. Sine is y divided by distance from the origin, with the sign of y.
  2. r²=8²+6²=100, so r=10.
  3. sin θ=6/10=3/5, positive in quadrant II.

3/5

WORKED EXAMPLE 2

The terminal ray of θ passes through (-15,8) in quadrant II. Find sin θ as a fraction.

  1. Sine is y divided by distance from the origin, with the sign of y.
  2. r²=15²+8²=289, so r=17.
  3. sin θ=8/17=8/17, positive in quadrant II.

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