HW4 Activity M Worksheet: Graphs of Sine, Cosine, and Tangent with shifts.

Let’s solve each problem step by step.

We are given trigonometric functions and asked to:
1. Find the amplitude and period.
2. Graph two periods of the function (we’ll describe how to graph it, since we can’t draw here).

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Reminder: For a function like

> y = A·sin(Bx) or y = A·cos(Bx)

- Amplitude = |A| → This is the height from the center line to the peak (or trough).
- Period = 2π / |B| → This is how long one full cycle takes.

For tangent:
> y = A·tan(Bx)

- Amplitude → Tangent doesn’t have amplitude in the same way — it goes to infinity. But sometimes problems still ask for “amplitude” as |A| even though technically it’s not bounded. We’ll note that.
- Period = π / |B|

Also, if there’s a negative sign in front (like -sin or -cos), it flips the graph upside down — but doesn’t change amplitude or period.

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## Problem 1: y = -sin(π/2 x)

This is sine with:
- A = -1 → Amplitude = |-1| = 1
- B = π/2 → Period = 2π / (π/2) = 2π × (2/π) = 4

Graphing tip: Start at (0,0), go down to min at x=1, back to zero at x=2, up to max at x=3, back to zero at x=4 → that’s one period. Do it again for second period (x=4 to x=8).

✔ Amplitude = 1, Period = 4

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## Problem 2: f(x) = 3 tan(2x)

Tangent function:
- A = 3 → Often called “amplitude” in worksheets, even though tan has no real amplitude → we’ll say 3
- B = 2 → Period = π / 2

Note: Tan has vertical asymptotes every half-period. So between 0 and π/2, it goes from 0 up to +∞, then jumps to -∞ and comes back to 0 at π/2? Wait — actually, standard tan(x) has period π, so tan(2x) has period π/2.

One period: from x=0 to x=π/2. But tan(2x) has asymptote when 2x = π/2 → x = π/4. So actually, one full cycle is from -π/4 to π/4? Let’s clarify:

Standard tan(x): period π, asymptotes at ±π/2, etc.

So tan(2x): compress horizontally by factor 2 → period = π/2, asymptotes at 2x = ±π/2 → x = ±π/4.

So one period is from -π/4 to π/4? Actually, better to think: from 0 to π/2, but with asymptote at π/4.

Actually, let’s define one period as interval between two consecutive asymptotes.

Asymptotes occur when 2x = π/2 + nπ → x = π/4 + nπ/2

So between x = -π/4 and x = π/4 → that’s one period? Length = π/2 → yes.

But for graphing, usually start at 0. From 0 to π/2: at x=0, tan(0)=0; as x→π/4⁻, tan(2x)→+∞; as x→π/4⁺, tan(2x)→-∞; at x=π/2, tan(π)=0.

Wait — that’s not right. tan(2*(π/2)) = tan(π) = 0. But between π/4 and π/2, it goes from -∞ to 0.

So one full cycle is from 0 to π/2? But it crosses zero at 0 and π/2, and has asymptote at π/4. That’s actually half a cycle? No — for tangent, one full cycle is between two asymptotes.

Better: The function repeats every π/2 units. So we can graph from 0 to π/2, but it will show only half the "S" shape? Actually, no — tan(2x) from 0 to π/2 covers one full period because:

At x=0: tan(0)=0
At x=π/8: tan(π/4)=1
At x=π/4: undefined (asymptote)
At x=3π/8: tan(3π/4)= -1
At x=π/2: tan(π)=0

So from 0 to π/2, it goes 0 → +∞ (jump) → -∞ → 0. That’s one full period? Actually, yes — because after π/2, it repeats: at x=π/2 + ε, same as x=ε.

So period = π/2.

And “amplitude” = 3 (vertical stretch).

✔ Amplitude = 3, Period = π/2

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## Problem 3: f(x) = cos(2x)

Cosine function:
- A = 1 → Amplitude = 1
- B = 2 → Period = 2π / 2 = π

Graph: Starts at max (1) at x=0, goes to 0 at x=π/4, min (-1) at x=π/2, back to 0 at x=3π/4, back to 1 at x=π → that’s one period. Repeat for second period (π to 2π).

✔ Amplitude = 1, Period = π

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## Problem 4: y = -tan(4πx)

Tangent:
- A = -1 → “Amplitude” = 1 (absolute value)
- B = 4π → Period = π / (4π) = 1/4

Asymptotes when 4πx = π/2 + nπ → x = (π/2 + nπ)/(4π) = (1/2 + n)/4 = 1/8 + n/4

So asymptotes at x = ..., -1/8, 1/8, 3/8, 5/8, ...

One period between x= -1/8 and x=1/8? Length = 1/4 → yes.

Or from 0 to 1/4: at x=0, tan(0)=0; as x→1/8⁻, tan(4πx)→+∞; as x→1/8⁺, →-∞; at x=1/4, tan(π)=0.

Negative sign flips it: so instead of going up to +∞ first, it goes down to -∞ first.

So graph: at x=0, y=0; as x increases to 1/8, y→-∞; then from 1/8 to 1/4, y goes from +∞ down to 0.

✔ Amplitude = 1, Period = 1/4

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## Problem 5: f(x) = -2 cos(π/3 x)

Cosine:
- A = -2 → Amplitude = |-2| = 2
- B = π/3 → Period = 2π / (π/3) = 2π × 3/π = 6

Negative sign flips the cosine wave upside down.

Normal cos starts at max, this starts at min.

At x=0: f(0) = -2 * cos(0) = -2 * 1 = -2
At x=3: cos(π/3 * 3) = cos(π) = -1 → f(3) = -2*(-1) = 2
At x=6: cos(2π) = 1 → f(6) = -2*1 = -2 → back to start.

So one period from 0 to 6.

✔ Amplitude = 2, Period = 6

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## Problem 6: y = sin(3/2 x)

Sine:
- A = 1 → Amplitude = 1
- B = 3/2 → Period = 2π / (3/2) = 2π × 2/3 = 4π/3

Graph: Standard sine shape, stretched horizontally.

Starts at (0,0), goes up to max at x = (1/4)*period = (1/4)*(4π/3) = π/3
Back to zero at x = π/2 * period? Wait — better:

Quarter points:

- 0: sin(0) = 0
- 1/4 period: x = (1/4)*(4π/3) = π/3 → sin(3/2 * π/3) = sin(π/2) = 1 → max
- 1/2 period: x = 2π/3 → sin(3/2 * 2π/3) = sin(π) = 0
- 3/4 period: x = π → sin(3/2 * π) = sin(3π/2) = -1 → min
- Full period: x = 4π/3 → sin(2π) = 0

Then repeat.

✔ Amplitude = 1, Period = 4π/3

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Now, compiling all answers:

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Final Answer:

1) Amplitude = 1, Period = 4
2) Amplitude = 3, Period = π/2
3) Amplitude = 1, Period = π
4) Amplitude = 1, Period = 1/4
5) Amplitude = 2, Period = 6
6) Amplitude = 1, Period = 4π/3