Session 04-04 - Tasks

Introduction to Trigonometric Functions

Trigonometric Functions - Problem Set

Problem 1: Angle Conversion (x)

Convert between degrees and radians:

Convert 225° to radians
Convert 7π/6 radians to degrees
Convert -120° to radians
Convert 3π/4 radians to degrees

Solution

Part a) 225° to radians:

Using the conversion factor: \(\frac{\pi \text{ radians}}{180°}\)

\[225° \times \frac{\pi}{180°} = \frac{225\pi}{180} = \frac{5\pi}{4}\]

Part b) 7π/6 radians to degrees:

Using the conversion factor: \(\frac{180°}{\pi \text{ radians}}\)

\[\frac{7\pi}{6} \times \frac{180°}{\pi} = \frac{7 \times 180°}{6} = \frac{1260°}{6} = 210°\]

Part c) -120° to radians:

\[-120° \times \frac{\pi}{180°} = \frac{-120\pi}{180} = \frac{-2\pi}{3}\]

Part d) 3π/4 radians to degrees:

\[\frac{3\pi}{4} \times \frac{180°}{\pi} = \frac{3 \times 180°}{4} = \frac{540°}{4} = 135°\]

Problem 2: Exact Trigonometric Values (x)

Find the exact values without a calculator:

sin(π/6)
cos(π/3)
tan(π/4)
sin(3π/2)

Solution

Part a) sin(π/6):

π/6 = 30°. From the special angles table: \[\sin(π/6) = \sin(30°) = \frac{1}{2}\]

Part b) cos(π/3):

π/3 = 60°. From the special angles table: \[\cos(π/3) = \cos(60°) = \frac{1}{2}\]

Part c) tan(π/4):

π/4 = 45°. From the special angles table: \[\tan(π/4) = \tan(45°) = 1\]

(Since sin(45°) = cos(45°) = √2/2, so tan(45°) = sin/cos = 1)

Part d) sin(3π/2):

3π/2 = 270°. This is at the bottom of the unit circle. \[\sin(3π/2) = \sin(270°) = -1\]

(The y-coordinate at the bottom of the unit circle is -1)

Problem 3: Unit Circle Coordinates (xx)

Find the exact coordinates (cos θ, sin θ) on the unit circle for:

θ = 2π/3
θ = 5π/4
θ = 11π/6
θ = 4π/3

Solution

Part a) θ = 2π/3 (120°):

Step 1: Identify the quadrant

2π/3 is between π/2 and π → Second quadrant

Step 2: Find the reference angle

Reference angle = π - 2π/3 = π/3 (60°)

Step 3: Apply quadrant signs

In quadrant II: cos is negative, sin is positive
cos(2π/3) = -cos(π/3) = -1/2
sin(2π/3) = sin(π/3) = √3/2

Coordinates: (-1/2, √3/2)

Part b) θ = 5π/4 (225°):

Step 1: Identify the quadrant

5π/4 is between π and 3π/2 → Third quadrant

Step 2: Find the reference angle

Reference angle = 5π/4 - π = π/4 (45°)

Step 3: Apply quadrant signs

In quadrant III: both cos and sin are negative
cos(5π/4) = -cos(π/4) = -√2/2
sin(5π/4) = -sin(π/4) = -√2/2

Coordinates: (-√2/2, -√2/2)

Part c) θ = 11π/6 (330°):

Step 1: Identify the quadrant

11π/6 is between 3π/2 and 2π → Fourth quadrant

Step 2: Find the reference angle

Reference angle = 2π - 11π/6 = π/6 (30°)

Step 3: Apply quadrant signs

In quadrant IV: cos is positive, sin is negative
cos(11π/6) = cos(π/6) = √3/2
sin(11π/6) = -sin(π/6) = -1/2

Coordinates: (√3/2, -1/2)

Part d) θ = 4π/3 (240°):

Step 1: Identify the quadrant

4π/3 is between π and 3π/2 → Third quadrant

Step 2: Find the reference angle

Reference angle = 4π/3 - π = π/3 (60°)

Step 3: Apply quadrant signs

In quadrant III: both cos and sin are negative
cos(4π/3) = -cos(π/3) = -1/2
sin(4π/3) = -sin(π/3) = -√3/2

Coordinates: (-1/2, -√3/2)

Problem 4: Analyzing Trigonometric Functions (xx)

For each function, find the amplitude, period, and range:

y = 2sin(x)
y = cos(3x)
y = -3sin(x/2)
y = 4cos(2x) + 1

Solution

Part a) y = 2sin(x):

Amplitude: |2| = 2
Period: 2π/1 = 2π
Range: [-2, 2]

Part b) y = cos(3x):

Amplitude: |1| = 1
Period: 2π/3
Range: [-1, 1]

Part c) y = -3sin(x/2):

Amplitude: |-3| = 3
Period: 2π/(1/2) = 4π
Range: [-3, 3]

Part d) y = 4cos(2x) + 1:

Amplitude: |4| = 4
Period: 2π/2 = π
Vertical shift: up 1 unit
Range: [1-4, 1+4] = [-3, 5]

Note: The range is affected by both amplitude and vertical shift.

Problem 5: Graphing Transformations (xx)

Sketch one period of each function and identify key features:

y = sin(x - π/4)
y = 2cos(x) - 1
y = sin(2x)
y = -cos(x + π/2)

Solution

Part a) y = sin(x - π/4):

Step 1: Identify the transformation

Horizontal shift right by π/4 (phase shift)
No amplitude change (A = 1)
No period change (period = 2π)

Step 2: Start with standard sine key points and shift each right by π/4:

(0, 0) → (π/4, 0)
(π/2, 1) → (3π/4, 1)
(π, 0) → (5π/4, 0)
(3π/2, -1) → (7π/4, -1)
(2π, 0) → (9π/4, 0)

Step 3: Plot these points and draw smooth sine curve

Range: [-1, 1]

Part b) y = 2cos(x) - 1:

Step 1: Identify transformations

Amplitude: |2| = 2 (vertical stretch)
Vertical shift: down 1 unit
Period: 2π (unchanged)

Step 2: Apply transformations to standard cosine:

Start with cos(x) key points
Multiply y-values by 2
Then subtract 1

Key points:

x = 0: y = 2(1) - 1 = 1 [maximum]
x = π/2: y = 2(0) - 1 = -1
x = π: y = 2(-1) - 1 = -3 [minimum]
x = 3π/2: y = 2(0) - 1 = -1
x = 2π: y = 2(1) - 1 = 1

Step 3: Plot and connect with smooth curve

Range: [-3, 1]

Part c) y = sin(2x):

Step 1: Identify the transformation

Period change: B = 2, so period = 2π/2 = π
Amplitude: 1 (unchanged)
Frequency doubles (completes cycle twice as fast)

Step 2: Find key points for one period [0, π]:

x = 0: y = sin(0) = 0
x = π/4: y = sin(π/2) = 1 [maximum]
x = π/2: y = sin(π) = 0
x = 3π/4: y = sin(3π/2) = -1 [minimum]
x = π: y = sin(2π) = 0

Step 3: Plot these points

The function completes a full cycle in π instead of 2π
Second cycle from π to 2π repeats the pattern

Part d) y = -cos(x + π/2):

Step 1: Identify transformations

Horizontal shift left by π/2
Reflection over x-axis (negative sign)

Step 2: Method 1 - Transform cosine:

Start with cos(x + π/2) key points (shift left π/2)
Then reflect over x-axis (multiply by -1)

Method 2 - Recognize the identity:

-cos(x + π/2) = sin(x)
This is a standard sine function!

Step 3: Key points (as sine function):

(0, 0)
(π/2, 1) [maximum]
(π, 0)
(3π/2, -1) [minimum]
(2π, 0)

Important: This demonstrates the co-function identity!

Problem 6: Ferris Wheel Application (xxx)

A Ferris wheel has a radius of 15 meters and its center is 18 meters above ground. It completes one rotation every 3 minutes, starting with a rider at the bottom.

Write a function h(t) for the height of a rider at time t minutes
What is the rider’s height after 45 seconds?
What is the rider’s height after 1 minute? After 1.5 minutes?
What is the maximum and minimum height?

Solution

Part a) Height function:

Setting up the model:

Center height: 18 meters
Radius (amplitude): 15 meters
Period: 3 minutes, so B = 2π/3
Starting at bottom means we use negative cosine

\[h(t) = 18 - 15\cos\left(\frac{2\pi t}{3}\right)\]

Alternative form using sine with phase shift: \[h(t) = 18 + 15\sin\left(\frac{2\pi t}{3} - \frac{\pi}{2}\right)\]

Part b) Height after 45 seconds:

Convert 45 seconds to minutes: t = 0.75 minutes

\[h(0.75) = 18 - 15\cos\left(\frac{2\pi \cdot 0.75}{3}\right)\] \[= 18 - 15\cos\left(\frac{\pi}{2}\right)\] \[= 18 - 15 \cdot 0 = 18\]

The rider is at 18 meters (level with the center).

Part c) Heights at 1 minute and 1.5 minutes:

At t = 1 minute: \[h(1) = 18 - 15\cos\left(\frac{2\pi \cdot 1}{3}\right)\] \[= 18 - 15\cos\left(\frac{2\pi}{3}\right)\] \[= 18 - 15 \cdot (-\frac{1}{2})\] \[= 18 + 7.5 = 25.5 \text{ meters}\]

At t = 1.5 minutes: \[h(1.5) = 18 - 15\cos\left(\frac{2\pi \cdot 1.5}{3}\right)\] \[= 18 - 15\cos(\pi)\] \[= 18 - 15 \cdot (-1)\] \[= 18 + 15 = 33 \text{ meters}\]

The rider is at 25.5 meters after 1 minute and at the maximum height of 33 meters after 1.5 minutes.

Part d) Maximum and minimum heights:

Maximum: 18 + 15 = 33 meters (at the top)
Minimum: 18 - 15 = 3 meters (at the bottom)

Problem 7: Temperature Modeling (xxxx)

The daily temperature in a city can be modeled by: \[T(h) = 20 + 8\sin\left(\frac{\pi}{12}(h - 6)\right)\]

where T is temperature in °C and h is the hour of the day (0 ≤ h ≤ 24).

What is the period of this function? What does this represent?
At what time does the maximum temperature occur? What is the maximum temperature?
At what time does the minimum temperature occur? What is the minimum temperature?
Find the temperature at 9:00 AM, noon, 3:00 PM, and 9:00 PM.

Solution

Part a) Period:

The function has the form \(T = A\sin(B(h - C)) + D\) where \(B = \frac{\pi}{12}\)

Period = \(\frac{2\pi}{B} = \frac{2\pi}{\pi/12} = 24\) hours

This represents one complete daily temperature cycle.

Part b) Maximum temperature:

The sine function reaches its maximum value of 1 when: \[\sin\left(\frac{\pi}{12}(h - 6)\right) = 1\]

This occurs when \(\frac{\pi}{12}(h - 6) = \frac{\pi}{2}\)

Solving: \(h - 6 = 6\), so \(h = 12\) (noon)

Maximum temperature: \(T(12) = 20 + 8(1) = 28°C\)

The maximum temperature of 28°C occurs at noon.

Part c) Minimum temperature:

The sine function reaches its minimum value of -1 when: \[\sin\left(\frac{\pi}{12}(h - 6)\right) = -1\]

This occurs when \(\frac{\pi}{12}(h - 6) = -\frac{\pi}{2}\) or \(\frac{3\pi}{2}\)

For the first occurrence: \(h - 6 = -6\), so \(h = 0\) (midnight)

Minimum temperature: \(T(0) = 20 + 8(-1) = 12°C\)

The minimum temperature of 12°C occurs at midnight.

Part d) Temperatures at specific times:

At 9:00 AM (h = 9): \[T(9) = 20 + 8\sin\left(\frac{\pi}{12}(9 - 6)\right) = 20 + 8\sin\left(\frac{\pi}{4}\right)\] \[= 20 + 8 \cdot \frac{\sqrt{2}}{2} = 20 + 4\sqrt{2} \approx 25.66°C\]

At noon (h = 12): \[T(12) = 28°C\] (from part b)

At 3:00 PM (h = 15): \[T(15) = 20 + 8\sin\left(\frac{\pi}{12}(15 - 6)\right) = 20 + 8\sin\left(\frac{3\pi}{4}\right)\] \[= 20 + 8 \cdot \frac{\sqrt{2}}{2} = 20 + 4\sqrt{2} \approx 25.66°C\]

At 9:00 PM (h = 21): \[T(21) = 20 + 8\sin\left(\frac{\pi}{12}(21 - 6)\right) = 20 + 8\sin\left(\frac{5\pi}{4}\right)\] \[= 20 + 8 \cdot (-\frac{\sqrt{2}}{2}) = 20 - 4\sqrt{2} \approx 14.34°C\]

Summary: The temperature rises from morning to noon, stays warm in the afternoon, then cools in the evening.