Homework worksheet on writing equations for sine and cosine functions with solved examples.

You're working on "Writing Equations for Sine and Cosine" — a standard trigonometry homework that asks you to write equations of the form:

> Sine: `y = a·sin(bx - c) + d`
> Cosine: `y = a·cos(bx - c) + d`

Where:
- `a` = amplitude (vertical stretch/compression, and sign indicates reflection)
- `b` = affects period: `period = 2π / |b|`
- `c` = horizontal shift (phase shift): `phase shift = c / b`
- `d` = vertical shift (midline)

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Let’s solve each problem step-by-step based on the graphs and handwritten work.

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## 🔹 PROBLEM 1

➤ Graph Analysis

- Amplitude (a): The graph goes from y = -3 to y = -1 → distance is 2 → half is 1 → a = 1
- Vertical Shift (d): Midline is at y = -2 → d = -2
- Period: One full cycle from x = -π/2 to x = π/2 → length = π → so period = π
- Then: `b = 2π / period = 2π / π = 2`
- Phase Shift (for sine): Standard sine starts at midline going up. Here, the first “up” point is at x = π/4 → so phase shift is π/4 to the right → `c/b = π/4` → since `b=2`, then `c = π/2`
- For cosine: Standard cosine starts at max. Here, the first max is at x = π/2 → so phase shift is π/2 to the right → `c/b = π/2` → `c = π`

✔ So:
- Sine equation: `y = 1·sin(2x - π/2) - 2` → or `y = sin(2x - π/2) - 2`
- Cosine equation: `y = 1·cos(2x - π) - 2` → or `y = cos(2x - π) - 2`

✔️ Matches what’s written in your worksheet!

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## 🔹 PROBLEM 2

➤ Graph Analysis

- Amplitude (a): Goes from y = -4 to y = 4 → amplitude = 4. But since it starts going *down* from origin, it’s reflected → a = -4 for sine, a = 4 for cosine (since cosine naturally starts at max, which matches this graph's start).
- Vertical Shift (d): Midline is y = 0 → d = 0
- Period: One cycle from x = 0 to x = 2π → period = 2π → `b = 2π / 2π = 1`
- Phase Shift (sine): Since sine normally starts at (0,0) going up, but here it starts going down → that’s equivalent to a reflection, so we can use `a = -4` and no phase shift → c = 0
- Phase Shift (cosine): Cosine normally starts at max. Here, the first max is at x = 3π/2 → so phase shift is 3π/2 to the right → `c/b = 3π/2` → since `b=1`, `c = 3π/2`

✔ So:
- Sine equation: `y = -4·sin(x) + 0` → `y = -4 sin(x)`
- Cosine equation: `y = 4·cos(x - 3π/2) + 0` → `y = 4 cos(x - 3π/2)`

✔️ Matches your worksheet!

> 💡 Note: You can also write `cos(x - 3π/2)` as `sin(x)` using identities, but the question asks for cosine form, so leave as is.

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## 🔹 PROBLEM 3 (Partially Visible — Let’s Reconstruct Based on What’s Shown)

From your handwritten notes:
- SINE: `a = -2` (upside down), `PS: left 2π` → phase shift left 2π
- COSINE: `a = 2`, `PS: left π`

Also visible: Period = 6π

Let’s verify and complete.

➤ Graph Analysis (Inferred)

- Amplitude (a): From graph (visible part), seems to go from ~y=0 to y=2 → amplitude = 2. But sine version is flipped → so a = -2 for sine, a = 2 for cosine
- Vertical Shift (d): Midline appears to be y = 1? Wait — let’s check.

Actually, looking at the graph (even though blurry), it looks like:
- Max ≈ y = 2
- Min ≈ y = 0
→ So midline = (2+0)/2 = 1 → d = 1

Wait — your notes say nothing about `d`. That might be an error. Let’s assume d = 1 based on graph.

- Period: Your note says `p = 6π` → so `b = 2π / 6π = 1/3`
- Phase Shift (sine): You wrote “left 2π” → so `c/b = -2π` → since `b = 1/3`, then `c = -2π * (1/3) = -2π/3`
- Phase Shift (cosine): You wrote “left π” → so `c/b = -π` → `c = -π * (1/3) = -π/3`

✔ So corrected equations:

- Sine: `y = -2·sin((1/3)x + 2π/3) + 1`
*(Note: phase shift left 2π → subtracting negative → adding positive inside)*
→ Actually: `y = -2 sin( (1/3)(x + 2π) ) + 1` → `y = -2 sin( (1/3)x + 2π/3 ) + 1`

- Cosine: `y = 2·cos( (1/3)x + π/3 ) + 1`
*(Phase shift left π → add π/3 inside)*

> 📝 Your worksheet may have forgotten the vertical shift `d=1`. That’s common! Always check if the graph is centered around y=0 or not.

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## ✔ FINAL ANSWERS:

Problem 1:

- Sine: `y = sin(2x - π/2) - 2`
- Cosine: `y = cos(2x - π) - 2`

Problem 2:

- Sine: `y = -4 sin(x)`
- Cosine: `y = 4 cos(x - 3π/2)`

Problem 3:

*(Assuming midline is y=1 based on graph)*
- Sine: `y = -2 sin( (1/3)x + 2π/3 ) + 1`
- Cosine: `y = 2 cos( (1/3)x + π/3 ) + 1`

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## 🧠 Tips for Future Problems:

1. Always find midline (d) first — average of max and min.
2. Amplitude (a) = half the distance between max and min. Sign depends on orientation.
3. Period = horizontal distance for one full cycle → `b = 2π / period`
4. Phase shift: For sine, find where the wave crosses midline going up. For cosine, find the first peak.
5. Use: `phase shift = c / b` → rearrange to get `c = b × phase shift`

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Let me know if you want to check Problem 3 with a clearer image or adjust for different assumptions!