This precalculus worksheet asks students to analyze trigonometric equations and sketch their graphs on provided coordinate planes.

Let’s solve each problem step by step. We’ll find:

- Amplitude
- Period (in radians)
- Phase shift (in radians)
- Vertical shift
- Minimum and maximum values

Then we’ll sketch the graph — but since we’re text-based, I’ll describe how to sketch it clearly so you can draw it yourself.

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Problem 1: y = 3 sin(θ - 5π/6)

This is in the form:
y = A sin(Bθ - C) + D
But here, B = 1, D = 0.

Actually, better to write as:
y = A sin(B(θ - φ)) + D
Where:
- A = amplitude
- Period = 2π / |B|
- φ = phase shift (right if positive)
- D = vertical shift

Our equation: y = 3 sin(θ - 5π/6)

So:
- A = 3 → Amplitude = 3
- B = 1 → Period = 2π / 1 = 2π
- Phase shift: inside the parentheses it’s (θ - 5π/6), so shift right by 5π/6
- D = 0 → Vertical shift = 0

Max value = D + A = 0 + 3 = 3
Min value = D - A = 0 - 3 = -3

To sketch:
- Start at θ = 5π/6 (phase shift), that’s where the sine wave starts its cycle (like sin(0)).
- At θ = 5π/6, y = 0 (since sin(0)=0)
- Then goes up to max at θ = 5π/6 + π/2 = 4π/3
- Back to zero at θ = 5π/6 + π = 11π/6
- Down to min at θ = 5π/6 + 3π/2 = 7π/3 (which is beyond 3π, so maybe stop at 3π)
- Back to zero at θ = 5π/6 + 2π = 17π/6 (also beyond 3π)

On your grid from 0 to 3π, plot key points starting at 5π/6.

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Problem 2: y = 3 cos θ

Standard cosine function scaled vertically.

Form: y = A cos(Bθ) + D

Here:
- A = 3 → Amplitude = 3
- B = 1 → Period = 2π
- No horizontal shift → Phase shift = 0
- D = 0 → Vertical shift = 0

Max = 3, Min = -3

Sketch:
- Starts at (0, 3) because cos(0) = 1 → 3*1 = 3
- Goes down to (π, -3)
- Back to (2π, 3)
- So on [0, 3π], it completes one full cycle by 2π, then half more to 3π (ends at -3)

Plot: (0,3), (π/2,0), (π,-3), (3π/2,0), (2π,3), (5π/2,0), (3π,-3)

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Problem 3: y = 2 sin(-3θ - π/2) + 2

First, simplify using identity: sin(-x) = -sin(x)

So:
y = 2 * [ -sin(3θ + π/2) ] + 2
→ y = -2 sin(3θ + π/2) + 2

Now write in standard form: y = A sin(B(θ - φ)) + D

Factor out B=3 inside:

y = -2 sin[ 3(θ + π/6) ] + 2

Because: 3θ + π/2 = 3(θ + π/6)

So now:

- A = -2 → but amplitude is absolute value → Amplitude = 2
- B = 3 → Period = 2π / 3
- Phase shift: θ + π/6 → shift left by π/6 → Phase shift = -π/6 (or “left π/6”)
- D = 2 → Vertical shift = 2

Max = D + |A| = 2 + 2 = 4
Min = D - |A| = 2 - 2 = 0

Note: The negative sign flips the sine wave upside down.

To sketch:
Start at phase shift: θ = -π/6 (but your graph starts at -π/4? Wait, looking at axis: x-axis goes from -π/4 to 2π. So -π/6 ≈ -0.52, which is between -π/4≈-0.78 and 0. So start there.

At θ = -π/6, argument of sin is 0 → sin(0)=0 → y = -2*(0)+2 = 2

But because of the negative sign, instead of going up, it goes DOWN first.

Key points for one period (length 2π/3):

Start: θ = -π/6 → y = 2
Quarter period later: θ = -π/6 + (2π/3)/4 = -π/6 + π/6 = 0 → sin(3*0 + π/2) = sin(π/2)=1 → y = -2*(1)+2 = 0
Half period: θ = -π/6 + π/3 = π/6 → sin(3*(π/6)+π/2)=sin(π/2 + π/2)=sin(π)=0 → y=2
Three-quarters: θ = -π/6 + π/2 = π/3 → sin(3*(π/3)+π/2)=sin(π + π/2)=sin(3π/2)=-1 → y = -2*(-1)+2 = 4
Full period: θ = -π/6 + 2π/3 = π/2 → sin(3*(π/2)+π/2)=sin(2π)=0 → y=2

So points: (-π/6, 2), (0, 0), (π/6, 2), (π/3, 4), (π/2, 2)

Repeat every 2π/3.

Your graph goes to 2π, so you’ll have about 3 full cycles.

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Problem 4: y = cos(θ/4 + π/4) - 2

Write as: y = cos( (1/4)(θ + π) ) - 2

Because: θ/4 + π/4 = (1/4)(θ + π)

So:

- A = 1 → Amplitude = 1
- B = 1/4 → Period = 2π / (1/4) = 8π
- Phase shift: θ + π → shift left by π → Phase shift = -π
- D = -2 → Vertical shift = -2

Max = -2 + 1 = -1
Min = -2 - 1 = -3

To sketch:

Start at θ = -π (phase shift). But your graph starts at θ=0? Looking at axis: from 0 to 12π.

So at θ=0:
y = cos(0/4 + π/4) - 2 = cos(π/4) - 2 = √2/2 - 2 ≈ 0.707 - 2 ≈ -1.293

But let’s use key points based on period 8π.

One full cycle from θ = -π to θ = -π + 8π = 7π

But since our graph starts at 0, let’s compute from θ=0 onward.

We know cosine starts at max when argument is 0.

Set argument = 0: θ/4 + π/4 = 0 → θ = -π → not in range.

Next key point: when argument = π/2 → cos=0

θ/4 + π/4 = π/2 → θ/4 = π/4 → θ = π → y = 0 - 2 = -2

Argument = π → cos=-1 → y = -1 -2 = -3

θ/4 + π/4 = π → θ/4 = 3π/4 → θ = 3π

Argument = 3π/2 → cos=0 → y=-2

θ/4 + π/4 = 3π/2 → θ/4 = 5π/4 → θ = 5π

Argument = 2π → cos=1 → y=1-2=-1

θ/4 + π/4 = 2π → θ/4 = 7π/4 → θ = 7π

So key points in [0, 12π]:

θ=0: y≈ -1.293 (as above)
θ=π: y=-2
θ=3π: y=-3 (min)
θ=5π: y=-2
θ=7π: y=-1 (max)
θ=9π: y=-2
θ=11π: y=-3
θ=12π: y = cos(12π/4 + π/4) -2 = cos(3π + π/4) -2 = cos(13π/4) -2
cos(13π/4) = cos(13π/4 - 2π*1) = cos(5π/4) = -√2/2 ≈ -0.707 → y≈ -2.707

But since period is 8π, from θ=0 to θ=8π is almost one full cycle? Let's see:

From θ=0 to θ=8π:

At θ=8π: argument = 8π/4 + π/4 = 2π + π/4 → cos(2π + π/4)=cos(π/4)=√2/2 → y≈ -1.293 same as θ=0

So yes, period 8π.

Plot points: (0, ~-1.3), (π, -2), (3π, -3), (5π, -2), (7π, -1), (8π, ~-1.3), etc.

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Final Answers Summary:

1) y = 3 sin(θ - 5π/6)
Amplitude: 3
Period: 2π
Phase shift: 5π/6 right
Vertical shift: 0
Max: 3, Min: -3

2) y = 3 cos θ
Amplitude: 3
Period: 2π
Phase shift: 0
Vertical shift: 0
Max: 3, Min: -3

3) y = 2 sin(-3θ - π/2) + 2
Amplitude: 2
Period: 2π/3
Phase shift: π/6 left (or -π/6)
Vertical shift: 2
Max: 4, Min: 0

4) y = cos(θ/4 + π/4) - 2
Amplitude: 1
Period: 8π
Phase shift: π left (or -π)
Vertical shift: -2
Max: -1, Min: -3

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Final Answer:
1) Amp=3, Per=2π, PS=5π/6 right, VS=0, Max=3, Min=-3
2) Amp=3, Per=2π, PS=0, VS=0, Max=3, Min=-3
3) Amp=2, Per=2π/3, PS=π/6 left, VS=2, Max=4, Min=0
4) Amp=1, Per=8π, PS=π left, VS=-2, Max=-1, Min=-3