3.8 trig identities WS Solutions .pdf - A 3-1 Solutions Name Date ... - Free Printable

Let’s solve each problem step by step. These are basic trig identities — we’re simplifying expressions using known relationships between sine, cosine, tangent, etc.

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Problem 1:
Simplify:
\[
\frac{\sin \theta}{\tan \theta} = \frac{\sin \theta}{?} = \boxed{?}
\]

We know that:
\[
\tan \theta = \frac{\sin \theta}{\cos \theta}
\]

So substitute that in the denominator:

\[
\frac{\sin \theta}{\tan \theta} = \frac{\sin \theta}{\frac{\sin \theta}{\cos \theta}} = \sin \theta \cdot \frac{\cos \theta}{\sin \theta} = \cos \theta
\]

✔ So first blank: \(\frac{\sin \theta}{\cos \theta}\) (which is tan θ), but since they wrote “= sinθ / ?”, the ? should be tan θ, and then simplified to cos θ

Wait — looking at the worksheet layout:

It says:
\[
\frac{\sin \theta}{\tan \theta} = \frac{\sin \theta}{?} = \boxed{?}
\]

Actually, it's probably meant to fill in the identity for tan θ in the denominator.

So:

Step 1: Replace tan θ with its definition → \(\frac{\sin \theta}{\cos \theta}\)

So:
\[
\frac{\sin \theta}{\tan \theta} = \frac{\sin \theta}{\frac{\sin \theta}{\cos \theta}} = \cos \theta
\]

But the way it’s written: “= sinθ / ?” — so the ? is tan θ, which equals \(\frac{\sin \theta}{\cos \theta}\), but maybe they want you to write what tan θ is?

Looking again — perhaps the first blank is where you write what tan θ equals, so:

\[
\frac{\sin \theta}{\tan \theta} = \frac{\sin \theta}{\frac{\sin \theta}{\cos \theta}} = \cos \theta
\]

So if the format is:

\[
\frac{\sin \theta}{\tan \theta} = \frac{\sin \theta}{\text{[blank]}} = \boxed{\text{[answer]}}
\]

Then [blank] = \(\frac{\sin \theta}{\cos \theta}\), and answer = \(\cos \theta\)

But that seems messy. Maybe they just want you to simplify directly.

Alternatively, maybe the first blank is “tan θ” itself? But that doesn’t help.

Wait — let me look at the second one for pattern.

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Problem 2:
\[
\cot \theta \cdot \sin \theta = \boxed{?}
\]

We know:
\[
\cot \theta = \frac{\cos \theta}{\sin \theta}
\]

So:

\[
\cot \theta \cdot \sin \theta = \frac{\cos \theta}{\sin \theta} \cdot \sin \theta = \cos \theta
\]

✔ Answer: \(\cos \theta\)

---

Back to Problem 1 — likely they want:

\[
\frac{\sin \theta}{\tan \theta} = \frac{\sin \theta}{\frac{\sin \theta}{\cos \theta}} = \cos \theta
\]

So if the blanks are:

First blank (denominator): \(\frac{\sin \theta}{\cos \theta}\)
Second blank (final answer): \(\cos \theta\)

But in the image, it looks like they have:

“= sinθ / ?” — so ? is tan θ, but then next step is boxed answer.

Actually, re-examining the original image description — user said “Substitution” worksheet.

Probably for #1:

They expect:

\[
\frac{\sin \theta}{\tan \theta} = \frac{\sin \theta}{\frac{\sin \theta}{\cos \theta}} = \cos \theta
\]

So if the first blank is under the fraction bar after “sinθ /”, it might be expecting you to write “tan θ” there, but that’s circular.

Wait — no, I think the structure is:

Left side: \(\frac{\sin \theta}{\tan \theta}\)

Then they show: = \(\frac{\sin \theta}{?}\) ← this ? is where you put the expression for tan θ, i.e., \(\frac{\sin \theta}{\cos \theta}\)

Then = boxed answer: \(\cos \theta\)

Yes, that makes sense.

So for #1:

First blank: \(\frac{\sin \theta}{\cos \theta}\)
Final answer: \(\cos \theta\)

But in the image, it might be written as two separate boxes.

Looking at your transcription:

You wrote:

> 1. sinθ / tanθ = sinθ / ? = □
> cotθ · sinθ = □

Ah! So actually, problem 1 has two parts? Or is it two separate problems?

In your text:

"1. sinθ / tanθ = sinθ / ? = □
cotθ · sinθ = □"

That suggests problem 1 is only the first line, and problem 2 is the second line.

But in standard worksheets, usually numbered separately.

Wait — in your initial message, you listed:

> 1. sinθ / tanθ = sinθ / ? = □
> 2. cotθ · sinθ = □
> 3. ... etc.

So yes, 8 problems total.

Let me list them clearly based on your transcription:

---

Problem 1:
\[
\frac{\sin \theta}{\tan \theta} = \frac{\sin \theta}{?} = \boxed{?}
\]

As above:

Replace tan θ with \(\frac{\sin \theta}{\cos \theta}\):

\[
\frac{\sin \theta}{\tan \theta} = \frac{\sin \theta}{\frac{\sin \theta}{\cos \theta}} = \cos \theta
\]

So the “?” in the denominator is \(\frac{\sin \theta}{\cos \theta}\), and final answer is \(\cos \theta\)

But perhaps they just want the final simplified form in the box, and the middle step is optional.

Given the format, I’ll assume:

For Problem 1:
The expression simplifies to \(\cos \theta\)

---

Problem 2:
\[
\cot \theta \cdot \sin \theta = \boxed{?}
\]

As above:
\(\cot \theta = \frac{\cos \theta}{\sin \theta}\), so multiplying by \(\sin \theta\) gives \(\cos \theta\)

✔ Answer: \(\cos \theta\)

---

Problem 3:
\[
\sec \theta - (\sec \theta \cos^2 \theta) = \boxed{?}
\]

Factor out sec θ:

\[
\sec \theta (1 - \cos^2 \theta)
\]

Recall identity: \(1 - \cos^2 \theta = \sin^2 \theta\)

So:

\[
\sec \theta \cdot \sin^2 \theta
\]

And \(\sec \theta = \frac{1}{\cos \theta}\), so:

\[
\frac{1}{\cos \theta} \cdot \sin^2 \theta = \frac{\sin^2 \theta}{\cos \theta}
\]

But can we simplify more? Not really, unless they want it in terms of other functions.

Wait — let me check the original image description — you had:

> 3. secθ - (secθ cos²θ) = □

And in your transcription, you wrote:

> = secθ(1 - cos²θ) = secθ(sin²θ) = ?

But then you have a box with “sin²θ / cosθ” or something?

Actually, in many cases, they might leave it as \(\sec \theta \sin^2 \theta\), but let’s see if it simplifies further.

Note: \(\sec \theta \sin^2 \theta = \frac{\sin^2 \theta}{\cos \theta} = \sin \theta \cdot \frac{\sin \theta}{\cos \theta} = \sin \theta \tan \theta\)

Is that simpler? Maybe.

But let’s see what’s expected.

Perhaps factor differently.

Another approach:

\[
\sec \theta - \sec \theta \cos^2 \theta = \sec \theta (1 - \cos^2 \theta) = \sec \theta \sin^2 \theta
\]

I think that’s acceptable, but let’s compute numerically to verify.

Suppose θ = 60°, so cos θ = 0.5, sin θ = √3/2 ≈ 0.866, sec θ = 2

Then: secθ - secθ cos²θ = 2 - 2*(0.25) = 2 - 0.5 = 1.5

Now, secθ sin²θ = 2 * (0.75) = 1.5 ✓

sinθ tanθ = (√3/2) * (√3) = (3/2) = 1.5 ✓

So both are correct, but perhaps they want it as \(\sin \theta \tan \theta\) or just \(\sec \theta \sin^2 \theta\)

Looking back at the worksheet, in problem 4, they have similar things.

Perhaps for consistency, we can write it as \(\sin \theta \tan \theta\), but let's hold on.

In your initial transcription, for problem 3, you have:

> = secθ(1 - cos²θ) = secθ(sin²θ) = □

And then you wrote "sin²θ / cosθ" which is the same as secθ sin²θ.

But in the box, perhaps they expect a single trig function or simplified form.

Notice that \(\frac{\sin^2 \theta}{\cos \theta} = \sin \theta \cdot \frac{\sin \theta}{\cos \theta} = \sin \theta \tan \theta\)

I think \(\sin \theta \tan \theta\) is a nice simplified form.

Let me confirm with another angle.

θ = 45°, cosθ = sinθ = √2/2 ≈ 0.707, secθ = √2 ≈ 1.414, tanθ = 1

Original: secθ - secθ cos²θ = 1.414 - 1.414*(0.5) = 1.414 - 0.707 = 0.707

sinθ tanθ = 0.707 * 1 = 0.707 ✓

secθ sin²θ = 1.414 * 0.5 = 0.707 ✓

So either is fine, but perhaps the worksheet expects \(\sin \theta \tan \theta\)

In problem 4, they have cscθ - cscθ sin²θ, which would be analogous.

Let's do problem 4 to see the pattern.

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Problem 4:
\[
\csc \theta - (\csc \theta \sin^2 \theta) = \boxed{?}
\]

Similarly:

Factor: \(\csc \theta (1 - \sin^2 \theta) = \csc \theta \cos^2 \theta\)

Since \(1 - \sin^2 \theta = \cos^2 \theta\)

And \(\csc \theta = \frac{1}{\sin \theta}\), so:

\[
\frac{\cos^2 \theta}{\sin \theta} = \cos \theta \cdot \frac{\cos \theta}{\sin \theta} = \cos \theta \cot \theta
\]

Numerical check: θ=60°, cscθ = 2/√3 ≈ 1.1547, sinθ=√3/2≈0.866, sin²θ=0.75

cscθ - cscθ sin²θ = 1.1547 - 1.1547*0.75 = 1.1547 - 0.866 = 0.2887

cosθ cotθ = 0.5 * (0.5 / 0.866) = 0.5 * 0.577 ≈ 0.2885 ✓

So for problem 3, similarly, it should be \(\sin \theta \tan \theta\)

For problem 4, \(\cos \theta \cot \theta\)

So I'll go with that.

Thus:

Problem 3: \(\sin \theta \tan \theta\)

Problem 4: \(\cos \theta \cot \theta\)

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Problem 5:
\[
\frac{\cos \theta}{\sec \theta} = \boxed{?}
\]

We know \(\sec \theta = \frac{1}{\cos \theta}\), so:

\[
\frac{\cos \theta}{\sec \theta} = \cos \theta \div \frac{1}{\cos \theta} = \cos \theta \cdot \cos \theta = \cos^2 \theta
\]

✔ Answer: \(\cos^2 \theta\)

---

Problem 6:
\[
\frac{\sin \theta}{\csc \theta} = \boxed{?}
\]

\(\csc \theta = \frac{1}{\sin \theta}\), so:

\[
\frac{\sin \theta}{\csc \theta} = \sin \theta \div \frac{1}{\sin \theta} = \sin \theta \cdot \sin \theta = \sin^2 \theta
\]

✔ Answer: \(\sin^2 \theta\)

---

Problem 7:
\[
\frac{\cos^2 \theta - \sin^2 \theta}{\cos^2 \theta} = \boxed{?}
\]

Split the fraction:

\[
\frac{\cos^2 \theta}{\cos^2 \theta} - \frac{\sin^2 \theta}{\cos^2 \theta} = 1 - \tan^2 \theta
\]

Because \(\frac{\sin^2 \theta}{\cos^2 \theta} = \tan^2 \theta\)

So answer: \(1 - \tan^2 \theta\)

We could also write it as \(\frac{\cos 2\theta}{\cos^2 \theta}\), but that's more complicated. The simplest is \(1 - \tan^2 \theta\)

Check with θ=45°: cos²θ = 0.5, sin²θ=0.5, so numerator=0, denominator=0.5, result=0

1 - tan²45° = 1 - 1 = 0 ✓

Good.

---

Problem 8:
\[
\frac{\sin \theta}{\cos \theta} \cdot \frac{\cos \theta}{\sin \theta} = \boxed{?}
\]

Multiply straight across:

Numerator: sinθ * cosθ
Denominator: cosθ * sinθ

So:

\[
\frac{\sin \theta \cos \theta}{\cos \theta \sin \theta} = 1
\]

Everything cancels.

✔ Answer: 1

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Now, let's compile all answers:

1. \(\frac{\sin \theta}{\tan \theta} = \cos \theta\)
(The intermediate step: replacing tanθ with sinθ/cosθ, but final answer is cosθ)

2. \(\cot \theta \cdot \sin \theta = \cos \theta\)

3. \(\sec \theta - \sec \theta \cos^2 \theta = \sin \theta \tan \theta\)

4. \(\csc \theta - \csc \theta \sin^2 \theta = \cos \theta \cot \theta\)

5. \(\frac{\cos \theta}{\sec \theta} = \cos^2 \theta\)

6. \(\frac{\sin \theta}{\csc \theta} = \sin^2 \theta\)

7. \(\frac{\cos^2 \theta - \sin^2 \theta}{\cos^2 \theta} = 1 - \tan^2 \theta\)

8. \(\frac{\sin \theta}{\cos \theta} \cdot \frac{\cos \theta}{\sin \theta} = 1\)

For problem 1, if they require filling the blank in "sinθ / ?", it should be \(\frac{\sin \theta}{\cos \theta}\), but since the final box is for the simplified answer, and in the context, I think for the final answer section, we put the simplified result.

Looking back at your transcription, for problem 1, you have:

> sinθ / tanθ = sinθ / ? = □

So the □ is the final simplified value, which is cosθ.

Similarly for others.

So I'll proceed.

Final Answers:

1. \(\cos \theta\)
2. \(\cos \theta\)
3. \(\sin \theta \tan \theta\)
4. \(\cos \theta \cot \theta\)
5. \(\cos^2 \theta\)
6. \(\sin^2 \theta\)
7. \(1 - \tan^2 \theta\)
8. \(1\)

Let me double-check problem 3 and 4.

For problem 3: secθ - secθ cos²θ = secθ(1 - cos²θ) = secθ sin²θ = (1/cosθ) * sin²θ = sin²θ / cosθ = sinθ * (sinθ / cosθ) = sinθ tanθ ✓

For problem 4: cscθ - cscθ sin²θ = cscθ(1 - sin²θ) = cscθ cos²θ = (1/sinθ) * cos²θ = cos²θ / sinθ = cosθ * (cosθ / sinθ) = cosθ cotθ ✓

All good.

Now, for the response, since ExplainAnswer is false, TeachTopic is false, GeneratePractice is false, I only need to provide the Final Answer section with all answers.

But the instruction says: "Complete ALL reasoning and explanations BEFORE writing the Final Answer." — but since those options are false, I don't include those sections. However, I did the reasoning in my thought process.

In the response, I should only output the Final Answer part, as per the rules.

The rules say: "If an option is false, completely omit that section." And "Always follow the structure below."

The structure has Explanation, TeachTopic, GeneratePractice, then Final Answer.

Since all are false, I omit the first three, and only write Final Answer.

But the Final Answer should contain the answers to all problems.

How to present? Probably list them as 1 to 8.

So:

Final Answer:
1. \(\cos \theta\)
2. \(\cos \theta\)
3. \(\sin \theta \tan \theta\)
4. \(\cos \theta \cot \theta\)
5. \(\cos^2 \theta\)
6. \(\sin^2 \theta\)
7. \(1 - \tan^2 \theta\)
8. \(1\)

To make it clear, perhaps box each or just list.

Since it's a worksheet, listing is fine.

I think that's it.