PDF) Trig Graphs Worksheet | Natasa Nikolovska Grujovska ... - Free Printable

It looks like you've uploaded two images of sinusoidal graphs, but the actual task or question isn't specified in your message. However, based on the content of the images, I can infer that you're likely asking for help analyzing these waveforms—possibly to determine their equations, compare them, or identify key features.

Let me analyze both graphs and explain what they represent.

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Graph 1 (Top Image):

This graph shows a periodic wave with the following characteristics:

- It appears to be a sine or cosine function.
- The wave has:
- A maximum value (amplitude) of about 2 (since it reaches up to 2 on the y-axis).
- A minimum value of about -2, so amplitude $ A = 2 $.
- The period is the distance between two consecutive peaks or troughs.
- From the x-axis labels: one peak is at $ x = \pi/2 $, the next at $ x = 5\pi/2 $.
- So period $ T = 5\pi/2 - \pi/2 = 2\pi $.
- Therefore, angular frequency $ B = \frac{2\pi}{T} = 1 $.
- The wave is symmetric about the origin and passes through zero at $ x = 0 $, suggesting it's a sine function (not cosine), since $ \sin(0) = 0 $.
- No horizontal shift (phase shift), and no vertical shift (midline is $ y = 0 $).

✔ So, the equation is likely:
$$
y = 2\sin(x)
$$

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Graph 2 (Bottom Image):

This graph also shows a sine-like wave, but let’s analyze it carefully:

- Amplitude: The maximum is about 1, minimum is -1, so $ A = 1 $.
- Period: From the x-axis:
- One peak is around $ x = \pi $, next at $ x = 3\pi $ → period $ T = 2\pi $
- So again, $ B = 1 $
- Midline: The wave oscillates around $ y = 0 $, so no vertical shift.
- At $ x = 0 $, the function value is negative (around $ y = -0.5 $), but not zero.
- But wait: it crosses the x-axis at $ x = 0 $, then goes down → this suggests it might be a negative sine or shifted.
- Actually, looking closely:
- At $ x = 0 $, $ y = 0 $
- Then it goes downward, so it's negative sine: $ y = -\sin(x) $
- Or possibly $ y = \sin(x - \phi) $

But let’s check:
- If $ y = \sin(x) $, it starts at 0 and goes up.
- Here, it starts at 0 and goes down, so it must be $ y = -\sin(x) $

Wait — but at $ x = \pi/2 $, $ y = -1 $? No, from the graph:
- At $ x = \pi/2 $, the function is at its minimum? No, look:
- The graph dips below zero after $ x = 0 $, reaches a minimum around $ x = \pi $, then rises back to zero at $ x = 2\pi $, and peaks at $ x = 3\pi $?

Wait — let’s recheck the x-axis labels:

From bottom graph:
- x-values: $ -\pi, -0.5\pi, 0, 0.5\pi, \pi, 1.5\pi, 2\pi, 2.5\pi $
- The wave:
- Crosses zero at $ x = 0 $
- Goes down to a minimum around $ x = \pi $
- Then comes back up to zero at $ x = 2\pi $
- Then goes up to a peak at $ x = 3\pi $

So the shape is:
- Starts at 0
- Goes down → minimum at $ x = \pi $
- Back to zero at $ x = 2\pi $
- Up to max at $ x = 3\pi $

That is exactly the shape of $ y = -\sin(x) $, because:
- $ \sin(0) = 0 $
- $ \sin(\pi) = 0 $, but derivative negative → going down → matches
- Wait, $ \sin(x) $ at $ x = \pi/2 $ is 1, but here the minimum is at $ x = \pi $? That doesn’t match.

Wait — actually, if the minimum is at $ x = \pi $, and amplitude is 1, then:
- $ y = -\cos(x) $? Let’s test:
- $ \cos(0) = 1 $ → $ y = -1 $
- But at $ x = 0 $, $ y = 0 $ → so not cosine.

Wait — perhaps it's a sine wave shifted?

Let’s suppose:
- The wave has a phase shift
- It crosses zero at $ x = 0 $, going downward
- Then reaches a minimum at $ x = \pi $
- Then back to zero at $ x = 2\pi $
- Then to a maximum at $ x = 3\pi $

This is exactly the behavior of $ y = -\sin(x) $, but wait:
- $ -\sin(x) $:
- $ x = 0 $: $ y = 0 $
- $ x = \pi/2 $: $ y = -1 $
- $ x = \pi $: $ y = 0 $
- $ x = 3\pi/2 $: $ y = 1 $
- $ x = 2\pi $: $ y = 0 $

But in the graph:
- Minimum occurs at $ x = \pi $, not $ x = \pi/2 $
- So that doesn’t match.

Wait — maybe the period is different?

Look at the x-axis:
- From $ x = 0 $ to $ x = 2\pi $: one full cycle?
- At $ x = 0 $: $ y = 0 $
- At $ x = \pi $: minimum
- At $ x = 2\pi $: $ y = 0 $
- At $ x = 3\pi $: maximum
- At $ x = 4\pi $: $ y = 0 $

Wait — the graph only shows up to $ x = 2.5\pi $, but we see:
- Zero at $ x = 0 $
- Min at $ x = \pi $
- Zero at $ x = 2\pi $
- Max at $ x = 3\pi $

So from $ x = 0 $ to $ x = 2\pi $: goes down and back up → half cycle?

Wait — no. From $ x = 0 $ to $ x = 2\pi $:
- Starts at 0, goes down to min at $ \pi $, back to 0 at $ 2\pi $
- Then from $ 2\pi $ to $ 3\pi $: goes up to max
- Then from $ 3\pi $ to $ 4\pi $: comes back down

So the full period is $ 4\pi $? Because it takes $ 4\pi $ to repeat?

Let’s check:
- Peak at $ x = 3\pi $, next peak should be at $ x = 7\pi $? But graph ends at $ 2.5\pi $, so we can't confirm.

Wait — but the first peak is at $ x = 3\pi $? And before that, at $ x = -\pi $, there's a peak?

Yes! Look at the left side:
- At $ x = -\pi $: a peak
- Then goes down to zero at $ x = 0 $
- Then to minimum at $ x = \pi $
- Then back to zero at $ x = 2\pi $
- Then up to peak at $ x = 3\pi $

So:
- Peak at $ x = -\pi $
- Next peak at $ x = 3\pi $
- Difference: $ 3\pi - (-\pi) = 4\pi $

So period $ T = 4\pi $

Then angular frequency $ B = \frac{2\pi}{T} = \frac{2\pi}{4\pi} = \frac{1}{2} $

Amplitude: from peak to midline is 1 → $ A = 1 $

Midline: $ y = 0 $

Now, phase shift: at $ x = -\pi $, we have a maximum

For a cosine function: $ y = A\cos(B(x - C)) $

We want maximum at $ x = -\pi $

Standard cosine: $ \cos(0) = 1 $ → maximum at $ x = C $

So set $ C = -\pi $

Thus:
$$
y = \cos\left(\frac{1}{2}(x + \pi)\right)
$$

Check:
- At $ x = -\pi $: $ y = \cos(0) = 1 $ ✔
- At $ x = 0 $: $ y = \cos(\frac{1}{2}(\pi)) = \cos(\pi/2) = 0 $ ✔
- At $ x = \pi $: $ y = \cos(\frac{1}{2}(2\pi)) = \cos(\pi) = -1 $ ✔
- At $ x = 2\pi $: $ y = \cos(\frac{1}{2}(3\pi)) = \cos(3\pi/2) = 0 $ ✔
- At $ x = 3\pi $: $ y = \cos(\frac{1}{2}(4\pi)) = \cos(2\pi) = 1 $ ✔

Perfect!

So the function is:
$$
y = \cos\left(\frac{1}{2}(x + \pi)\right)
$$

Alternatively, simplify:
$$
y = \cos\left(\frac{x}{2} + \frac{\pi}{2}\right)
$$

And using identity: $ \cos(\theta + \pi/2) = -\sin(\theta) $, so:
$$
y = -\sin\left(\frac{x}{2}\right)
$$

✔ So the equation is $ y = -\sin\left(\frac{x}{2}\right) $

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Summary of Both Graphs:

1. Top Graph:
- Equation: $ y = 2\sin(x) $
- Amplitude: 2
- Period: $ 2\pi $
- Phase shift: 0
- Vertical shift: 0

2. Bottom Graph:
- Equation: $ y = -\sin\left(\frac{x}{2}\right) $
- Amplitude: 1
- Period: $ 4\pi $
- Phase shift: none (or equivalent to $ \pi/2 $ in cosine form)
- Vertical shift: 0

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Possible Task?

If the task was to write equations for both graphs, here are the answers:

- Top: $ \boxed{y = 2\sin(x)} $
- Bottom: $ \boxed{y = -\sin\left(\frac{x}{2}\right)} $

If the task was something else (e.g., find amplitude, period, phase shift, etc.), please clarify, and I can adjust accordingly.

Let me know if you need further explanation or comparison!