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๐ŸŽฏโญ INTERACTIVE LESSON

Graphing Trigonometric Functions

Learn step-by-step with interactive practice!

Graphing Trigonometric Functions - Complete Interactive Lesson

Part 1: Sine & Cosine Graphs

๐Ÿ“ Sine & Cosine Graphs

Part 1 of 7 โ€” Sine & Cosine Graphs

y=Asinโก(Bx+C)+Dy = A\sin(Bx + C) + D

  • A: amplitude (vertical stretch)
  • B: affects period (T=2ฯ€โˆฃBโˆฃT = \frac{2\pi}{|B|})
  • C: phase shift
  • D: vertical shift (midline)

Sine starts at midline; cosine starts at maximum.

Worked Example

y=sinโก(x)y = \sin(x). Key features?

Amplitude = 1, Period = 2ฯ€, Midline: y = 0 โœ…

Concept Check ๐ŸŽฏ

Amplitude ๐Ÿงฎ

  1. Amplitude of sinโก(x)\sin(x)?

  2. Amplitude of cosโก(x)\cos(x)?

  3. Amplitude of 2sinโก(x)2\sin(x)?

Concept Check ๐Ÿ”

Practice

#FunctionAmplitudePeriod
1sin(x)12ฯ€
2cos(x)12ฯ€
32sin(x)22ฯ€

Challenge Question ๐Ÿ“‹

Part 2: Amplitude & Period

๐Ÿ“Š Amplitude & Period

Part 2 of 7 โ€” Amplitude & Period

  • Amplitude = โˆฃAโˆฃ|A| (distance from midline to max/min)
  • Period = 2ฯ€โˆฃBโˆฃ\frac{2\pi}{|B|} (one full cycle length)

For y=3sinโก(2x)y = 3\sin(2x): amplitude = 3, period = 2ฯ€2=ฯ€\frac{2\pi}{2} = \pi

Worked Example

y=4cosโก(3x)y = 4\cos(3x). Amplitude? Period?

Amplitude = 4, Period = 2ฯ€3โ‰ˆ2.09\frac{2\pi}{3} \approx 2.09 โœ…

Concept Check ๐ŸŽฏ

Find Amplitude ๐Ÿงฎ

  1. 3sinโก(2x)3\sin(2x): Amplitude?

  2. 5cosโก(x)5\cos(x): Amplitude?

  3. sinโก(4x)\sin(4x): Amplitude?

Concept Check ๐Ÿ”

Practice

#FunctionAmplitudePeriod
13sinโก(2x)3\sin(2x)3ฯ€
25cosโก(x)5\cos(x)52ฯ€
3sinโก(4x)\sin(4x)1ฯ€/2

Challenge Question ๐Ÿ“‹

Part 3: Phase Shift

๐Ÿ”ข Phase Shift

Part 3 of 7 โ€” Phase Shift

y=Asinโก(B(xโˆ’h))+Dy = A\sin(B(x - h)) + D

Phase shift = h=โˆ’CBh = -\frac{C}{B} (horizontal shift)

  • Shift right when h>0h > 0
  • Shift left when h<0h < 0

Vertical shift = DD (moves the midline)

Worked Example

y=sinโก(xโˆ’ฯ€4)y = \sin(x - \frac{\pi}{4}). Phase shift?

Shift right by ฯ€4\frac{\pi}{4} โœ…

Concept Check ๐ŸŽฏ

Shifts ๐Ÿงฎ

  1. sinโก(x)+3\sin(x) + 3. Midline at y=y = ?

  2. sinโก(xโˆ’ฯ€)\sin(x - \pi). Midline at y=y = ?

  3. cosโก(x)+2\cos(x) + 2. Midline at y=y = ?

Concept Check ๐Ÿ”

Practice

#FunctionPhase Shift
1sinโก(xโˆ’ฯ€)\sin(x - \pi)Right ฯ€
2cosโก(x+ฯ€2)\cos(x + \frac{\pi}{2})Left ฯ€/2
3sinโก(x)+3\sin(x) + 3No phase shift, up 3

Challenge Question ๐Ÿ“‹

Part 4: Modeling with Sinusoids

๐Ÿ“ˆ Modeling with Sinusoids

Part 4 of 7 โ€” Modeling with Sinusoids

Many real-world phenomena are periodic:

  • Temperature over a year
  • Tides, daylight hours
  • Sound waves, pendulums

To model: find amplitude (half of maxโˆ’min), midline (average of max and min), and period (time for one cycle).

Worked Example

Temperature: max 90ยฐF, min 30ยฐF, period 12 months.

A=90โˆ’302=30A = \frac{90-30}{2} = 30, D=90+302=60D = \frac{90+30}{2} = 60 B=2ฯ€12=ฯ€6B = \frac{2\pi}{12} = \frac{\pi}{6}

T(t)=30sinโก(ฯ€6t)+60T(t) = 30\sin\left(\frac{\pi}{6}t\right) + 60 โœ…

Concept Check ๐ŸŽฏ

Find Amplitude ๐Ÿงฎ

Amplitude = (max โˆ’ min) / 2

  1. Max=80, min=40. Amplitude?

  2. Max=100, min=20. Amplitude?

  3. Max=50, min=10. Amplitude?

Concept Check ๐Ÿ”

Practice

#MaxMinAmplitudeMidline
180402060
2100204060
350102030

Challenge Question ๐Ÿ“‹

Part 5: Inverse Trig Functions

๐Ÿงฎ Inverse Trig Functions

Part 5 of 7 โ€” Inverse Trig Functions

Sine, cosine and tangent are periodic, so each output comes from infinitely many angles. To build an inverse, the domain of each function is restricted to one interval where it takes every output exactly once. That interval becomes the range of the inverse function, and AP Precalculus measures it in radians.

Inverse functionInputOutput (range)Quadrants of the output
sinโกโˆ’1(x)\sin^{-1}(x) or arcsinโก(x)\arcsin(x)โˆ’1โ‰คxโ‰ค1-1 \le x \le 1[โˆ’ฯ€2,ฯ€2]\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]I and IV
cosโกโˆ’1(x)\cos^{-1}(x) or arccosโก(x)\arccos(x)โˆ’1โ‰คxโ‰ค1-1 \le x \le 1[0,ฯ€][0, \pi]I and II
tanโกโˆ’1(x)\tan^{-1}(x) or arctanโก(x)\arctan(x)all real xx(โˆ’ฯ€2,ฯ€2)\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)I and IV

๐Ÿ”‘ Key Concept: arcsinโก(x)\arcsin(x) is the angle in [โˆ’ฯ€2,ฯ€2]\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] whose sine is xx. A negative input gives a negative angle for arcsine and arctangent, but an angle between ฯ€2\frac{\pi}{2} and ฯ€\pi for arccosine.

Worked Examples

1) arcsinโก(โˆ’22)\arcsin\left(-\frac{\sqrt{2}}{2}\right). The reference angle with sine 22\frac{\sqrt{2}}{2} is ฯ€4\frac{\pi}{4}. The input is negative and the range is [โˆ’ฯ€2,ฯ€2]\left[-\frac{\pi}{2}, \frac{\pi}{2}\right], so the answer is in Quadrant IV: โˆ’ฯ€4-\frac{\pi}{4}.

2) arccosโก(โˆ’12)\arccos\left(-\frac{1}{2}\right). The reference angle with cosine 12\frac{1}{2} is ฯ€3\frac{\pi}{3}. The range [0,ฯ€][0, \pi] puts a negative cosine in Quadrant II: ฯ€โˆ’ฯ€3=2ฯ€3\pi - \frac{\pi}{3} = \frac{2\pi}{3}. The angle 4ฯ€3\frac{4\pi}{3} also has cosine โˆ’12-\frac{1}{2}, but it is outside the range.

3) arctanโก(โˆ’3)\arctan(-\sqrt{3}). The reference angle with tangent 3\sqrt{3} is ฯ€3\frac{\pi}{3}. A negative input gives a Quadrant IV angle: โˆ’ฯ€3-\frac{\pi}{3}.

ExpressionReference angleAnswer
arcsinโก(1)\arcsin(1)ฯ€2\frac{\pi}{2}ฯ€2\frac{\pi}{2}
arccosโก(0)\arccos(0)ฯ€2\frac{\pi}{2}ฯ€2\frac{\pi}{2}
arctanโก(1)\arctan(1)ฯ€4\frac{\pi}{4}ฯ€4\frac{\pi}{4}
arccosโก(โˆ’1)\arccos(-1)00ฯ€\pi

Concept Check ๐ŸŽฏ

Inverse Trig in Radians ๐Ÿงฎ

Give each answer in radians as a fraction of pi, such as "pi/3", "-pi/4" or "0".

  1. arccosโก(1)=\arccos(1) = ?

  2. arctanโก(โˆ’1)=\arctan(-1) = ?

  3. arccosโก(โˆ’32)=\arccos\left(-\frac{\sqrt{3}}{2}\right) = ?

Ranges in Radians ๐Ÿ”

Using an Inverse to Solve an Equation

An inverse function gives one solution. The other solutions come from the unit circle and the period.

Solve sinโกx=0.4\sin x = 0.4 for 0โ‰คx<2ฯ€0 \le x < 2\pi.

StepWork
One solution from the inversex=arcsinโก(0.4)โ‰ˆ0.412x = \arcsin(0.4) \approx 0.412, in Quadrant I
Second solution with the same sinex=ฯ€โˆ’arcsinโก(0.4)โ‰ˆ2.730x = \pi - \arcsin(0.4) \approx 2.730, in Quadrant II
Check the intervalBoth are between 00 and 2ฯ€2\pi; adding 2ฯ€2\pi would leave the interval

In a model such as h(t)=10+4sinโก(ฯ€6t)h(t) = 10 + 4\sin\left(\frac{\pi}{6}t\right), the same idea finds every time tt when h(t)h(t) reaches a given value: solve for the sine, apply the inverse, then use symmetry and the period to list the other times.

Challenge Question ๐Ÿ“‹

Part 6: Problem-Solving Workshop

๐Ÿ› ๏ธ Problem-Solving Workshop

Part 6 of 7 โ€” Problem-Solving Workshop

Apply all trig modeling concepts:

  • Read amplitude, period, shifts from a graph or description
  • Write a sinusoidal model
  • Evaluate inverse trig expressions

Worked Example

Ferris wheel: radius 20 ft, center 25 ft high, period 60 sec.

h(t)=โˆ’20cosโก(2ฯ€60t)+25=โˆ’20cosโก(ฯ€30t)+25h(t) = -20\cos\left(\frac{2\pi}{60}t\right) + 25 = -20\cos\left(\frac{\pi}{30}t\right) + 25 โœ…

(Negative cosine starts at the bottom)

Concept Check ๐ŸŽฏ

Applications ๐Ÿงฎ

Max = 70, min = 30:

  1. Midline?

  2. Amplitude?

  3. Amplitude of 5sinโก(ฯ€x)+25\sin(\pi x) + 2?

Concept Check ๐Ÿ”

Practice

#ProblemType
1Max 70, min 30, period 12Model
2arctan(0)Inverse trig
3Amplitude of 5sin(ฯ€x)+2Read graph

Challenge Question ๐Ÿ“‹

Part 7: Review & Applications

๐Ÿ† Review & Applications

Part 7 of 7 โ€” Review & Applications

Key Formulas

  • y=Asinโก(B(xโˆ’h))+Dy = A\sin(B(x-h)) + D
  • Amplitude = โˆฃAโˆฃ|A|
  • Period = 2ฯ€/โˆฃBโˆฃ2\pi / |B|
  • Phase shift = hh
  • Midline = DD
  • Amplitude = (maxโˆ’min)/2
  • Midline = (max+min)/2

Worked Example

y=3sinโก(2x)+1y = 3\sin(2x) + 1

A=3, B=2, D=1. Amplitude=3, Period=ฯ€, Midline: y=1 โœ…

Concept Check ๐ŸŽฏ

Review ๐Ÿงฎ

  1. Amplitude of 4cosโก(x)4\cos(x)?

  2. sinโกโˆ’1(1)\sin^{-1}(1) in radians? (Enter like "pi/3")

  3. Midline of sinโก(x)+1\sin(x) + 1?

Concept Check ๐Ÿ”

Practice

#TopicProblem
1Amplitudey=4cosโก(x)y = 4\cos(x)
2Period of sinโก(3x)\sin(3x)2ฯ€/3
3arcsin(1)ฯ€2\frac{\pi}{2}

Challenge Question ๐Ÿ“‹