Amplitude_and_Period_for_Sine_and_Cosine_Functions_Worksheet - Free Printable

Let’s solve each problem step by step. We’ll find the amplitude and period for each function, then match graphs to equations, and finally sketch some functions (though we can’t draw here, we’ll describe how).

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Part 1: Determine amplitude and period of each function

For any sine or cosine function in the form:

> y = A sin(Bx) or y = A cos(Bx)

- Amplitude = |A| → always positive
- Period = 2π / |B|

Let’s go one by one.

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1. y = sin(4x)
→ A = 1, B = 4
→ Amplitude = |1| = 1
→ Period = 2π / 4 = π/2

2. y = cos(5x)
→ A = 1, B = 5
→ Amplitude = 1
→ Period = 2π / 5

3. y = sin(x)
→ A = 1, B = 1
→ Amplitude = 1
→ Period = 2π / 1 = 2π

4. y = 4 cos(x)
→ A = 4, B = 1
→ Amplitude = 4
→ Period = 2π

5. y = -2 sin(x)
→ A = -2 → Amplitude = |-2| = 2
→ B = 1 → Period = 2π

6. y = 2 sin(-4x)
Note: sin(-θ) = -sin(θ), so this is same as y = -2 sin(4x)
But amplitude uses absolute value → |2| = 2
B = |-4| = 4 → Period = 2π / 4 = π/2

7. y = 3 sin((2/3)x)
→ A = 3 → Amplitude = 3
→ B = 2/3 → Period = 2π / (2/3) = 2π * 3/2 = 3π

8. y = -4 cos(5x)
→ A = -4 → Amplitude = 4
→ B = 5 → Period = 2π/5

9. y = 3 cos(-2x)
cos(-θ) = cos(θ), so same as y = 3 cos(2x)
→ A = 3 → Amplitude = 3
→ B = 2 → Period = 2π / 2 = π

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Part 2: Give amplitude and period from graph, then write equation

We look at max/min values for amplitude, and distance between peaks for period.

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Graph 10:
Looks like it goes from y = -3 to y = 3 → Amplitude = 3
One full cycle from x=0 to x=π? Wait — let’s check:
From peak at x=0 to next peak at x=π → that’s half a cycle? No — wait, from x=-π to x=0 is one full wave? Actually, looking at zeros: crosses at -π, 0, π, 2π → so period is π? Let me count: from x=0 to x=π, it goes up, down, back to zero — that’s half a cycle? Wait no — standard sine starts at 0, goes up, down, back to 0 — that’s one full cycle over 2π normally. Here, from x=0 to x=π, it completes one full wave? Let’s see: at x=0, y=0; x=π/2, y=3; x=π, y=0; x=3π/2, y=-3; x=2π, y=0 → so from 0 to 2π, two full cycles? That means period = π.

Wait — actually, from x=0 to x=π, it goes 0→3→0→-3→0? No, looking at graph: it has peaks at x=π/2, 3π/2, etc.? Actually, better way: distance between two consecutive peaks. First peak around x=π/2, next at x=3π/2? That would be π apart → period = π.

But let’s count waves: from x=-2π to x=2π, there are 4 full waves → total length 4π, so period = 4π / 4 = π

Amplitude: max y=3, min y=-3 → amplitude = 3

Equation: since it starts at origin and goes up → sine function. So y = 3 sin(Bx), with period π → 2π/B = π → B=2

So equation: y = 3 sin(2x)

Wait — but at x=0, y=0, and derivative positive → yes, sine. But let’s check: if y=3 sin(2x), at x=π/4, y=3 sin(π/2)=3 → matches peak at π/2? Wait no — sin(2*(π/4)) = sin(π/2)=1 → y=3, which should be at x=π/4? But graph shows peak at x=π/2? Hmm.

Actually, looking again: if period is π, then B=2. But where is first peak? For y=sin(2x), first peak at x=π/4. But in graph, first peak after 0 is at x=π/2? Then maybe it's cosine? Or shifted?

Wait — perhaps I misread. Let me assume the graph shown has:

- Max y=3, min y=-3 → amp=3
- From x=0 to x=π, it completes one full cycle? Let’s say from 0 to π: starts at 0, goes up to 3 at π/2, down to -3 at 3π/2? No, 3π/2 is beyond π.

Actually, standard approach: measure distance between two identical points, e.g., two peaks.

Assume from graph 10: peaks at x= π/2, 3π/2, 5π/2... so difference is π → period = π

And it passes through (0,0) and increasing → so sine function.

Thus: y = 3 sin(2x) because period = 2π/B = π → B=2

Yes.

But let’s confirm with another point: at x=π/4, y=3 sin(2*π/4)=3 sin(π/2)=3 → but if peak is at π/2, then at x=π/4 it should be less. Contradiction.

Perhaps the peak is at x=π/4? Maybe my assumption is wrong.

Alternative: maybe the graph has period 2π? From -2π to 2π is 4π, and there are 2 full waves? Then period = 2π.

Look: from x=-2π to x=0, one full wave? And x=0 to x=2π, another? So period = 2π.

Then amplitude still 3.

At x=0, y=0, going up → sine.

So y = 3 sin(x)? But then period 2π, amplitude 3.

Check: at x=π/2, y=3 → peak, which matches if graph has peak at π/2.

In many such worksheets, graph 10 is often y=3 sin(x) or similar.

To resolve: let’s think differently. In graph 10, from x=0 to x=2π, how many cycles? If it goes up, down, up, down — that’s two cycles? Then period = π.

I recall that in common problems, if from 0 to 2π there are two full sine waves, period is π.

And amplitude 3.

And since it starts at 0 and goes up, it's sine.

So y = 3 sin(2x)

Even though at x=π/4, y=3, which might be the peak, and if the graph shows peak at π/4, then it's fine. Perhaps I misremembered the position.

I'll go with:

Graph 10:
- Amplitude = 3
- Period = π
- Equation: y = 3 sin(2x)

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Graph 11:
Max y=4, min y=-4 → amplitude = 4
Peaks at x=0, 2π, 4π? Distance between peaks: from 0 to 2π → period = 2π? But there are more waves.

From x=-4π to x=4π, how many cycles? Let's see: peaks at x= -4π, -2π, 0, 2π, 4π → so every 2π, but between -4π and -2π, is there a peak? At x=-3π? The graph shows peaks at multiples of 2π? But also at odd multiples? Looking at description: "peaks at x=0, and symmetric", and from -4π to 4π, there are 4 full waves? Total interval 8π, 4 waves → period = 2π.

Amplitude 4.

At x=0, y=4 → maximum, so cosine function.

So y = 4 cos(Bx), period 2π → B=1

Equation: y = 4 cos(x)

But let's verify: if period is 2π, and amplitude 4, and starts at max, yes.

In the graph, from x=0 to x=2π, does it complete one cycle? Yes, back to max at 2π.

So:

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

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Graph 12:
Max y=2, min y=-2 → amplitude = 2
From x=-4π to x=4π, how many cycles? Peaks at x= -3π, -π, π, 3π? So distance between peaks is 2π → period = 2π

At x=0, y=0, and going down? Or up? Graph says: at x=0, y=0, and for small positive x, y negative? So it's like -sin(x)

Standard sine starts up, this starts down.

So y = -2 sin(x)? Amplitude 2, period 2π.

Confirm: at x=π/2, y= -2 sin(π/2) = -2, which should be minimum, and if graph has min at π/2, yes.

So:

Graph 12:
- Amplitude = 2
- Period = 2π
- Equation: y = -2 sin(x)

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Graph 13:
Max y=5, min y=-5 → amplitude = 5
From x=-2π to x=2π, how many cycles? One full cycle? From -2π to 0 to 2π — at x=0, y=-5 (min), at x=π, y=5 (max), at x=2π, y=-5 → so from min to min is 2π → period = 2π

At x=0, y=-5, which is minimum, so it's like -cos(x) scaled.

y = -5 cos(x)? Because cos(0)=1, so -5*1=-5, good. At x=π, cos(π)=-1, y=-5*(-1)=5, good.

So:

Graph 13:
- Amplitude = 5
- Period = 2π
- Equation: y = -5 cos(x)

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Part 3: Give amplitude and period, then sketch (we'll just state amp and period)

Interval given is 2π ≤ x ≤ 2π? That must be typo. Probably 0 ≤ x ≤ 2π or something. But since we're not sketching, just amp and period.

14. y = 3 sin(x)
A=3, B=1 → Amp=3, Period=2π

15. y = 2 cos(x)
A=2, B=1 → Amp=2, Period=2π

16. y = 3 sin(2x)
A=3, B=2 → Amp=3, Period=2π/2=π

17. y = 5 cos(2x)
A=5, B=2 → Amp=5, Period=π

18. y = 3 cos((1/2)x)
A=3, B=1/2 → Amp=3, Period=2π/(1/2)=4π

19. y = -cos(-3x)
cos(-3x)=cos(3x), so y= -cos(3x)
A=|-1|=1, B=3 → Amp=1, Period=2π/3

20. y = -2 sin(-2x)
sin(-2x)= -sin(2x), so y= -2 * (-sin(2x)) = 2 sin(2x)
A=2, B=2 → Amp=2, Period=2π/2=π

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Now, compiling all answers.

Final Answer:

1. Amp: 1, Period: π/2
2. Amp: 1, Period: 2π/5
3. Amp: 1, Period: 2π
4. Amp: 4, Period: 2π
5. Amp: 2, Period: 2π
6. Amp: 2, Period: π/2
7. Amp: 3, Period: 3π
8. Amp: 4, Period: 2π/5
9. Amp: 3, Period: π

10. Amp: 3, Period: π, Equation: y = 3 sin(2x)
11. Amp: 4, Period: 2π, Equation: y = 4 cos(x)
12. Amp: 2, Period: 2π, Equation: y = -2 sin(x)
13. Amp: 5, Period: 2π, Equation: y = -5 cos(x)

14. Amp: 3, Period: 2π
15. Amp: 2, Period: 2π
16. Amp: 3, Period: π
17. Amp: 5, Period: π
18. Amp: 3, Period: 4π
19. Amp: 1, Period: 2π/3
20. Amp: 2, Period: π