Handwritten notes on graphing sine and cosine functions, including amplitude, period, and phase shifts.

Let’s solve each of the 8 graphing problems one by one. We’ll find:

- The amplitude (a)
- The period
- The trig pattern (key points in one cycle)
- The exciting spots (where max, min, or midline crossings happen — often at quarter-periods)

We are told to graph between 0 and 2π unless otherwise specified.

---

1) y = cos b(x - h) + k → but no equation given? Wait — looking back, problem 1 is missing the actual function!

Wait — actually, looking at the image again, it seems like for #1, they wrote “y = cos b...” but didn’t specify values. But then below it says “Trig pattern: (0,1,-1,0,1)” — that suggests it's just y = cos x, since cosine starts at 1, goes to 0, -1, 0, 1 over [0, 2π].

Also, they wrote period = π/2? That doesn’t match cos x. Hmm.

Actually, let me re-read carefully.

Looking at the right side:

> Graph the function between 0 and 2π
> 1) y = cos b...
> Trig pattern: (0,1,-1,0,1) ← this looks wrong for cosine. Cosine should be (1,0,-1,0,1). Maybe typo?

But then they write:
period: π/2 = π? That math is wrong. π/2 ≠ π.

This seems messy. Let me ignore their handwritten notes and do it correctly from scratch.

Actually — wait! Looking more closely at the bottom right corner, there’s a box:

“Graph from 0 to 2π:
1) y = cos x
2) y = sin x
3) y = -cos x
4) y = -sin x
5) y = 2 sin 4x
6) y = -3 cos ½ x
7) y = -2 sin ½ x
8) y = ¼ cos 2x”

Ah! So those are the actual functions. The ones on top were probably examples or miswritten.

So we’ll use these 8 functions.

Let’s go one by one.

---

Problem 1: y = cos x

Standard cosine function.

- Amplitude: |a| = 1
- Period: 2π / b = 2π / 1 = 2π
- Trig pattern (one cycle from 0 to 2π):
At x=0 → y=1
x=π/2 → y=0
x=π → y=-1
x=3π/2 → y=0
x=2π → y=1
So pattern: (0,1), (π/2,0), (π,-1), (3π/2,0), (2π,1)

Exciting spots: same as above — peaks, troughs, zeros.

Period = 2π

Amplitude = 1

---

Problem 2: y = sin x

Standard sine.

- Amplitude: 1
- Period: 2π
- Trig pattern:
x=0 → 0
x=π/2 → 1
x=π → 0
x=3π/2 → -1
x=2π → 0
Pattern: (0,0), (π/2,1), (π,0), (3π/2,-1), (2π,0)

Exciting spots: same.

---

Problem 3: y = -cos x

Negative cosine — flips upside down.

- Amplitude: |-1| = 1
- Period: 2π
- Starts at -1 instead of 1.
- Pattern:
x=0 → -1
x=π/2 → 0
x=π → 1
x=3π/2 → 0
x=2π → -1
So: (0,-1), (π/2,0), (π,1), (3π/2,0), (2π,-1)

Exciting spots: same x-values, flipped y.

---

Problem 4: y = -sin x

Flipped sine.

- Amplitude: 1
- Period: 2π
- Pattern:
x=0 → 0
x=π/2 → -1
x=π → 0
x=3π/2 → 1
x=2π → 0
So: (0,0), (π/2,-1), (π,0), (3π/2,1), (2π,0)

---

Problem 5: y = 2 sin 4x

Now we have transformations.

General form: y = a sin(bx - h) + k → here h=0, k=0

- Amplitude: |a| = |2| = 2
- Period: 2π / b = 2π / 4 = π/2
- Since period is π/2, in [0, 2π] we will have 2π ÷ (π/2) = 4 full cycles.

Trig pattern for one cycle (from 0 to π/2):

Sine starts at 0, up to max, back to 0, down to min, back to 0.

Divide period into 4 parts: each part = (π/2)/4 = π/8

So key x-values for one cycle:
0, π/8, π/4, 3π/8, π/2

Corresponding y-values for sin(4x):

At x=0 → sin(0)=0 → y=0
x=π/8 → sin(4*π/8)=sin(π/2)=1 → y=2*1=2
x=π/4 → sin(π)=0 → y=0
x=3π/8 → sin(3π/2)=-1 → y=2*(-1)=-2
x=π/2 → sin(2π)=0 → y=0

So one cycle pattern: (0,0), (π/8,2), (π/4,0), (3π/8,-2), (π/2,0)

Then repeat every π/2.

Exciting spots: all those x-values where something happens — max, min, zero.

In [0, 2π], we’ll have 4 cycles → so 5 sets of these points? Actually, endpoints overlap.

Total exciting spots: list all x where y=0, ±2 within [0,2π]

Since period is π/2, and 4 cycles, we can generate:

Cycle 1: x=0, π/8, π/4, 3π/8, π/2
Cycle 2: x=π/2, 5π/8, 3π/4, 7π/8, π
Cycle 3: x=π, 9π/8, 5π/4, 11π/8, 3π/2
Cycle 4: x=3π/2, 13π/8, 7π/4, 15π/8, 2π

But note: x=π/2, π, 3π/2, 2π appear twice — we only list once.

So unique exciting spots x-values:
0, π/8, π/4, 3π/8, π/2, 5π/8, 3π/4, 7π/8, π, 9π/8, 5π/4, 11π/8, 3π/2, 13π/8, 7π/4, 15π/8, 2π

That’s 17 points? But maybe they just want the pattern per cycle and say how many cycles.

For simplicity, we can say:

Period: π/2
Amplitude: 2
One cycle pattern: (0,0), (π/8,2), (π/4,0), (3π/8,-2), (π/2,0)
Number of cycles in [0,2π]: 4
Exciting spots: occur every π/8 starting at 0, alternating max/min/zeros.

---

Problem 6: y = -3 cos(½ x)

Form: y = a cos(bx) + k, with a=-3, b=1/2, k=0

- Amplitude: |a| = 3
- Period: 2π / b = 2π / (1/2) = 4π
- But we’re only graphing from 0 to 2π → which is HALF a period!

Because full period is 4π, so from 0 to 2π is half-cycle.

Cosine normally: starts at max, goes to min at half-period.

Here, because of negative sign, it starts at MINIMUM.

Normal cos(½ x) at x=0: cos(0)=1 → y=-3*1=-3
At x=2π: cos(½ * 2π)=cos(π)=-1 → y=-3*(-1)=3
At x=π: cos(½ * π)=cos(π/2)=0 → y=0

So in [0,2π]:

x=0 → y=-3
x=π → y=0
x=2π → y=3

And since it’s cosine, smooth curve.

Pattern for half-cycle: start at min, go through midline at quarter-period? Wait.

Full period is 4π, so quarter-period is π.

From 0 to 4π:

- x=0: cos(0)=1 → y=-3
- x=π: cos(π/2)=0 → y=0
- x=2π: cos(π)=-1 → y=3
- x=3π: cos(3π/2)=0 → y=0
- x=4π: cos(2π)=1 → y=-3

But we only go to 2π, so we have first three points: (0,-3), (π,0), (2π,3)

Exciting spots: at x=0 (min), x=π (midline crossing), x=2π (max)

Period: 4π (but we don’t complete it)

Amplitude: 3

---

Problem 7: y = -2 sin(½ x)

a = -2, b = 1/2

- Amplitude: 2
- Period: 2π / (1/2) = 4π
- Again, graphing only to 2π → half period.

Sin(½ x) normally: starts at 0, goes up to max at π, back to 0 at 2π, etc.

But with negative sign: flips vertically.

So:

x=0: sin(0)=0 → y=0
x=π: sin(π/2)=1 → y=-2*1=-2
x=2π: sin(π)=0 → y=0

So points: (0,0), (π,-2), (2π,0)

It goes down to minimum at x=π.

Exciting spots: x=0 (start), x=π (min), x=2π (back to zero)

Period: 4π

Amplitude: 2

---

Problem 8: y = ¼ cos(2x)

a = 1/4, b=2

- Amplitude: |1/4| = 0.25
- Period: 2π / 2 = π
- In [0,2π], number of cycles: 2π / π = 2 full cycles.

Cos(2x) pattern per cycle (period π):

Divide π into 4 parts: each π/4

x=0: cos(0)=1 → y=0.25
x=π/4: cos(π/2)=0 → y=0
x=π/2: cos(π)=-1 → y=-0.25
x=3π/4: cos(3π/2)=0 → y=0
x=π: cos(2π)=1 → y=0.25

Then repeats for next cycle: x=π to 2π same as 0 to π.

So exciting spots: every π/4 from 0 to 2π.

List: 0, π/4, π/2, 3π/4, π, 5π/4, 3π/2, 7π/4, 2π

With corresponding y: 0.25, 0, -0.25, 0, 0.25, 0, -0.25, 0, 0.25

---

Now, to summarize answers clearly for each problem:

I think the student needs to fill in for each:

- a (amplitude, including sign if needed for direction)
- period
- trig pattern (for one cycle)
- exciting spots (x-values or points)

But since the original had some blanks, I’ll provide complete info for each.

Also, note: in the image, for #1 they had “period: π/2 = π?” which is incorrect — we fixed it.

Let me compile final answers per problem.

---

Final Answers:

1) y = cos x
a = 1
period = 2π
trig pattern: (0,1), (π/2,0), (π,-1), (3π/2,0), (2π,1)
exciting spots: x = 0, π/2, π, 3π/2, 2π

2) y = sin x
a = 1
period = 2π
trig pattern: (0,0), (π/2,1), (π,0), (3π/2,-1), (2π,0)
exciting spots: x = 0, π/2, π, 3π/2, 2π

3) y = -cos x
a = -1
period = 2π
trig pattern: (0,-1), (π/2,0), (π,1), (3π/2,0), (2π,-1)
exciting spots: x = 0, π/2, π, 3π/2, 2π

4) y = -sin x
a = -1
period = 2π
trig pattern: (0,0), (π/2,-1), (π,0), (3π/2,1), (2π,0)
exciting spots: x = 0, π/2, π, 3π/2, 2π

5) y = 2 sin 4x
a = 2
period = π/2
trig pattern (one cycle): (0,0), (π/8,2), (π/4,0), (3π/8,-2), (π/2,0)
exciting spots: x = 0, π/8, π/4, 3π/8, π/2, 5π/8, 3π/4, 7π/8, π, 9π/8, 5π/4, 11π/8, 3π/2, 13π/8, 7π/4, 15π/8, 2π
(or simply: every π/8 from 0 to 2π)

6) y = -3 cos(½ x)
a = -3
period = 4π
trig pattern (for half cycle in [0,2π]): (0,-3), (π,0), (2π,3)
exciting spots: x = 0, π, 2π

7) y = -2 sin(½ x)
a = -2
period = 4π
trig pattern (half cycle): (0,0), (π,-2), (2π,0)
exciting spots: x = 0, π, 2π

8) y = ¼ cos(2x)
a = 1/4
period = π
trig pattern (one cycle): (0,0.25), (π/4,0), (π/2,-0.25), (3π/4,0), (π,0.25)
exciting spots: x = 0, π/4, π/2, 3π/4, π, 5π/4, 3π/2, 7π/4, 2π

---

Note: For problems 6 and 7, since the period is 4π and we're only going to 2π, we don't complete a full cycle, so the "trig pattern" is only for the portion shown.

Also, "exciting spots" typically mean the key points: max, min, and where it crosses midline — which we've included.

I think this covers everything accurately.

Final Answer: See summarized answers above for each of the 8 functions.