Trigonometry worksheet focusing on sine and cosine function graphs, including amplitude, period, and equation writing.

Let’s go through each problem one by one. We’ll use the standard rules for sine and cosine functions:

For any function in the form:
- y = A sin(Bx) or y = A cos(Bx)

→ Amplitude = |A| (absolute value of A — always positive!)
→ Period = 2π / |B|

Note: The negative sign in front of B doesn’t affect period or amplitude — only the absolute value matters.

---

Problem 6: y = 2 sin(-4x)

Amplitude = |2| = 2

Period = 2π / |-4| = 2π / 4 = π/2

✔ Final Answer for #6:
Amplitude = 2
Period = π/2

---

Problem 9: y = 3 cos(-2x)

Amplitude = |3| = 3

Period = 2π / |-2| = 2π / 2 = π

✔ Final Answer for #9:
Amplitude = 3
Period = π

---

Now let’s look at the graphs.

---

Problem 10: Graph with x from -2π to 2π, peaks at y=3 and y=-3

First, find amplitude:
The graph goes from top (max) = 3 to bottom (min) = -3 → so amplitude is half the distance between max and min:
(3 - (-3)) / 2 = 6/2 = 3

Now, period:
Look at how long it takes to repeat. From x=0 to x=π, we see two full cycles? Wait — let’s count carefully.

From x=0 to x=π, there are two complete waves. So one wave happens over π/2.

Wait — actually, from x=0 to x=π, you can see: starts at 0, up to peak, down to trough, back to 0 — that’s ONE full cycle? No — wait, looking again:

Actually, from x=0 to x=π, the graph completes two full cycles. Let me check:

At x=0 → y=0
Then up to peak at ~π/4, down to trough at ~3π/4, back to 0 at π → that’s one full cycle? But then it repeats again before 2π.

Wait — better way: count how many cycles fit in 2π.

From x=0 to x=2π, how many full waves? Looking at the graph: it looks like 4 full cycles between 0 and 2π.

So period = 2π / 4 = π/2

But wait — let’s double-check with key points.

From x=0 to next time it returns to same point (same height and direction):
At x=0, y=0, going up. Next time it does that is at x=π/2? Let’s see:

If period is π/2, then yes — every π/2 units, it repeats.

Yes — because from 0 to π/2: one full wave. Then π/2 to π: second, etc.

So period = π/2

Now, what kind of function? It passes through origin and goes up first → that’s a sine function.

Equation: y = A sin(Bx)

We have A = 3, period = π/2 → so B = 2π / period = 2π / (π/2) = 4

So equation: y = 3 sin(4x)

✔ Final Answer for #10:
Amplitude = 3
Period = π/2
Equation: y = 3 sin(4x)

---

Problem 11: Graph with x from -4π to 4π, peaks at y=4 and y=-4

Amplitude: max = 4, min = -4 → amplitude = (4 - (-4))/2 = 8/2 = 4

Period: How many cycles between 0 and 2π?

Looking at graph: from x=0 to x=2π, how many full waves?

It looks like 2 full cycles between 0 and 2π → so period = 2π / 2 = π

Check: from x=0 to x=π, one full cycle? Yes — starts at peak, goes down to trough, back to peak → that’s one cycle for cosine.

Wait — at x=0, y=4 (peak) → that suggests it’s a cosine function (since cos(0)=1).

So equation: y = A cos(Bx)

A = 4

Period = π → so B = 2π / period = 2π / π = 2

Equation: y = 4 cos(2x)

✔ Final Answer for #11:
Amplitude = 4
Period = π
Equation: y = 4 cos(2x)

---

## Now compiling all answers:

For problems 6 and 9 (fill-in-the-blank):

6.
Amplitude = 2
Period = π/2

9.
Amplitude = 3
Period = π

For graph problems:

10.
Amplitude = 3
Period = π/2
Equation: y = 3 sin(4x)

11.
Amplitude = 4
Period = π
Equation: y = 4 cos(2x)

---

Final Answer:

6. Amplitude = 2, Period = π/2
9. Amplitude = 3, Period = π
10. Amplitude = 3, Period = π/2, Equation: y = 3 sin(4x)
11. Amplitude = 4, Period = π, Equation: y = 4 cos(2x)