Worksheet for graphing sine, cosine, and tangent functions with equations and blank grids for plotting.

Let’s solve each problem step by step. We’ll find the domain, range, amplitude, period, phase shift, and vertical slide for each trig function.

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General reminders:

For functions like:
- y = A sin(B(θ - C)) + D
- y = A cos(B(θ - C)) + D

Where:
- Amplitude = |A|
- Period = 2π / |B|
- Phase shift = C (to the right if positive, left if negative)
- Vertical slide (shift) = D
- Domain of sine/cosine is always all real numbers → (-∞, ∞) or “all real θ”
- Range depends on amplitude and vertical shift: [D - |A|, D + |A|]

Tangent has different rules — but none of these are tangent, so we’re safe.

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Problem 1: y = sin(2θ)

This is same as: y = 1·sin(2(θ - 0)) + 0

→ A = 1, B = 2, C = 0, D = 0

- Domain: All real numbers → (-∞, ∞) or just write “all real θ”
- Range: Since amplitude = 1, no vertical shift → [-1, 1]
- Amplitude: |1| = 1
- Period: 2π / 2 = π
- Phase shift: 0
- Vertical slide: 0

✔ Check: sin(2θ) oscillates between -1 and 1, repeats every π units, starts at origin.

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Problem 2: y = cos(½ θ)

Same as: y = 1·cos((1/2)(θ - 0)) + 0

→ A = 1, B = 1/2, C = 0, D = 0

- Domain: All real θ
- Range: [-1, 1] (amplitude 1, no vertical shift)
- Amplitude: 1
- Period: 2π / (1/2) = 4π
- Phase shift: 0
- Vertical slide: 0

✔ Check: cosine normally has period 2π; here it’s stretched horizontally → takes twice as long to repeat → 4π. Correct.

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Problem 3: y = cos(3θ)

Same as: y = 1·cos(3(θ - 0)) + 0

→ A = 1, B = 3, C = 0, D = 0

- Domain: All real θ
- Range: [-1, 1]
- Amplitude: 1
- Period: 2π / 3
- Phase shift: 0
- Vertical slide: 0

✔ Check: compressed horizontally → repeats faster → period smaller than 2π → 2π/3 is correct.

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Problem 4: y = sin(2(θ + π))

First, rewrite to match standard form:
y = sin(2(θ - (-π))) → so C = -π

→ A = 1, B = 2, C = -π, D = 0

- Domain: All real θ
- Range: [-1, 1]
- Amplitude: 1
- Period: 2π / 2 = π
- Phase shift: -π → which means shift LEFT by π
- But note: sometimes people say “phase shift = -π”, meaning left π. That’s acceptable.
- Alternatively, you can think: sin(2(θ + π)) = sin(2θ + 2π) = sin(2θ) because sine has period 2π → so actually, this is identical to sin(2θ)! Wait — let’s check that.

Wait — important!

sin(2(θ + π)) = sin(2θ + 2π)

But sin(x + 2π) = sin(x), so sin(2θ + 2π) = sin(2θ)

So technically, this function is IDENTICAL to y = sin(2θ). So phase shift should be 0? But according to formula, we have a shift of -π.

What’s going on?

Actually, when we write y = sin(2(θ + π)), the phase shift is calculated as: from θ - C, so θ + π = θ - (-π) → C = -π → phase shift = -π (left π).

BUT since the period is π, shifting left by π is the same as shifting by one full period → so graph looks identical.

In terms of answering the question, we go by the formula unless told otherwise.

So:

- Phase shift = -π (or “π to the left”)
- Vertical slide = 0

However, some teachers might accept 0 because it's equivalent due to periodicity. But strictly speaking, based on the expression given, phase shift is -π.

Let me double-check with an example:

At θ = 0:
Original sin(2θ) → sin(0) = 0
Our function: sin(2(0 + π)) = sin(2π) = 0 → same

At θ = π/4:
sin(2*(π/4)) = sin(π/2) = 1
Our function: sin(2*(π/4 + π)) = sin(2*(5π/4)) = sin(5π/2) = sin(π/2) = 1 → same

So yes, they are identical. Therefore, even though algebraically we get phase shift = -π, visually and functionally, it’s the same as no shift.

But in most textbooks, when asked for phase shift from the equation y = sin(2(θ + π)), they expect you to report the value from the formula: phase shift = -π.

To avoid confusion, I’ll follow the strict interpretation: use the formula.

So:

- Phase shift: -π (or “π units to the left”)
- But note: since period is π, shifting by π brings you back to start → so graph unchanged.

Still, for homework purposes, we'll put what the formula gives.

Alternatively, maybe simplify first:

y = sin(2(θ + π)) = sin(2θ + 2π) = sin(2θ) → so effectively, phase shift = 0.

Hmm. This is tricky.

Let me see how standard sources handle this.

Actually, the definition of phase shift is the horizontal displacement from the parent function. Since sin(2θ + 2π) = sin(2θ), there is NO actual displacement — it’s the same graph.

Therefore, the phase shift should be 0.

But why does the formula give -π?

Because the formula assumes you don’t simplify using periodic identities. In reality, phase shift is defined modulo the period.

So best answer: since sin(2(θ + π)) = sin(2θ), the phase shift is 0.

I think for educational purposes, especially at high school level, if simplifying shows it’s identical to original, then phase shift is 0.

Let me confirm with another point:

Parent function sin(2θ): zero at θ=0, max at θ=π/4, etc.

Our function: sin(2(θ+π)): at θ = -π, we get sin(2(-π + π)) = sin(0) = 0 → so the "starting point" is shifted to θ=-π.

But since the function repeats every π, having a zero at θ=-π is the same as having one at θ=0, because -π + π = 0.

So again, it’s indistinguishable.

Given that, I think the expected answer is to treat it as written without simplifying via periodicity — so phase shift = -π.

But let’s look at common textbook practice.

Upon reflection: many curricula teach to extract phase shift directly from the equation without simplifying using periodic properties. So for y = sin(2(θ + π)), they would say phase shift is -π.

Moreover, if we didn't know about periodicity, we'd still calculate it that way.

So to stay consistent with typical classroom expectations, I’ll go with:

Phase shift = -π

But note: in reality, it doesn’t change the graph.

Final decision: Use formula-based answer.

So:

- Domain: all real θ
- Range: [-1, 1]
- Amplitude: 1
- Period: π
- Phase shift: -π (or “π to the left”)
- Vertical slide: 0

Some may argue for 0, but I’ll stick with -π as per direct extraction.

Actually, wait — let’s write it clearly:

The general form is y = A sin(B(θ - C)) + D

We have y = sin(2(θ + π)) = sin(2(θ - (-π)))

So C = -π → phase shift = C = -π

Yes.

So final answers:

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## Final Answer:

1) y = sin(2θ)
Domain: all real θ
Range: [-1, 1]
Amplitude: 1
Period: π
Phase shift: 0
Vertical slide: 0

2) y = cos(½ θ)
Domain: all real θ
Range: [-1, 1]
Amplitude: 1
Period: 4π
Phase shift: 0
Vertical slide: 0

3) y = cos(3θ)
Domain: all real θ
Range: [-1, 1]
Amplitude: 1
Period: 2π/3
Phase shift: 0
Vertical slide: 0

4) y = sin(2(θ + π))
Domain: all real θ
Range: [-1, 1]
Amplitude: 1
Period: π
Phase shift: -π (or π to the left)
Vertical slide: 0

Note: For #4, although the phase shift is mathematically -π, the graph is identical to y = sin(2θ) due to the periodic nature of sine. However, based on the given equation form, the phase shift is reported as -π.