Writing equations for sine and cosine graphs with visual examples and formulas.

Let’s solve this step by step.

We are given a graph of a sine or cosine function, and we need to write its equation. The general form for these functions is:

> y = A·sin(B(x - C)) + D
> or
> y = A·cos(B(x - C)) + D

Where:
- A = amplitude (how tall the wave is from center to peak)
- B = affects period → Period = 2π / B
- C = horizontal shift (phase shift)
- D = vertical shift (where the midline is)

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Step 1: Find the vertical shift (D)

Look at the graph — the middle line (midline) of the wave is halfway between the highest and lowest points.

From the image:
- Highest point (max) = 18
- Lowest point (min) = 2

So, midline = (18 + 2) / 2 = 20 / 2 = 10

→ So, D = 10

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Step 2: Find the amplitude (A)

Amplitude = distance from midline to max (or min)

= 18 - 10 = 8

→ So, A = 8

(Note: If the graph starts going down from the midline, we might use negative amplitude — we’ll check that later.)

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Step 3: Find the period

Period = length of one full cycle.

Looking at the graph:
- One full wave goes from x = 4 to x = 20? Let’s check.

Actually, look at two consecutive peaks:
- First peak around x = 8
- Next peak around x = 24

Wait — let’s use the labeled points.

In the handwritten notes, it says “Per = 16” — that matches if you go from x=4 to x=20 (that’s 16 units), or from x=8 to x=24.

Yes — so Period = 16

Now, since Period = 2π / B → then B = 2π / Period = 2π / 16 = π/8

But in the answer box, they wrote 2π/16 — which simplifies to π/8. Both are correct, but 2π/16 is kept as-is sometimes for clarity.

→ So, B = 2π/16 (we can leave it like that for now)

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Step 4: Find horizontal shift (C)

This depends on whether we’re using sine or cosine.

#### Option 1: Using Sine

Sine normally starts at the midline and goes up.

On our graph, where does it cross the midline going upward?

Looking at the graph: At x = 4, the graph crosses the midline (y=10) and is going UP.

That’s exactly where sine starts!

So for sine: y = A·sin(B(x - C)) + D

We want the “start” point to be at x = 4 → so C = 4

Thus:
y = 8·sin( (2π/16)(x - 4) ) + 10

Which matches the first equation in the pink box.

#### Option 2: Using Cosine

Cosine normally starts at a maximum.

Our graph has a maximum at x = 8.

So if we use cosine, we shift right by 8.

Thus:
y = 8·cos( (2π/16)(x - 8) ) + 10

Matches second equation.

#### Option 3: Negative Cosine

What if we start at a minimum? Our graph has a minimum at x = 0? Wait — at x=0, y=2? Actually, looking at the graph, at x=0, it's at the bottom? But wait — there’s a circle at x=0, y=2 — yes, that’s a minimum.

Cosine normally starts at max. If we flip it (negative), it starts at min.

So if we use negative cosine, and no shift? Let’s see:

At x=0, we have a minimum → perfect for -cos(0) = -1 → times amplitude 8 → -8, plus 10 → 2. Yes!

So:
y = -8·cos( (2π/16)x ) + 10

Because at x=0: cos(0)=1 → -8*1 +10 = 2 → correct.

And at x=8: cos( (2π/16)*8 ) = cos(π) = -1 → -8*(-1)+10 = 8+10=18 → correct.

So all three equations are valid representations of the same graph.

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Final Check:

All three equations should give the same values at key points.

Let’s test x=4:

1. y = 8·sin( (2π/16)(4-4) ) +10 = 8·sin(0) +10 = 0 +10 = 10 ✔ (midline, going up)

2. y = 8·cos( (2π/16)(4-8) ) +10 = 8·cos( (2π/16)(-4) ) +10 = 8·cos(-π/2) +10 = 8·0 +10 = 10 ✔

3. y = -8·cos( (2π/16)*4 ) +10 = -8·cos(π/2) +10 = -8·0 +10 = 10 ✔

Good.

Test x=8:

1. sin( (2π/16)(8-4) ) = sin( (2π/16)*4 ) = sin(π/2) = 1 → 8*1+10=18 ✔

2. cos( (2π/16)(8-8) ) = cos(0) = 1 → 8*1+10=18 ✔

3. -8·cos( (2π/16)*8 ) = -8·cos(π) = -8*(-1)=8 → 8+10=18 ✔

Perfect.

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Final Answer:
The three possible equations for the graph are:
y = 8 sin( (2π/16)(x - 4) ) + 10
y = 8 cos( (2π/16)(x - 8) ) + 10
y = -8 cos( (2π/16)x ) + 10