Use identities to find the indicated value for each angle measure.

Find 

sine left parenthesis 2 theta right parenthesissin(2θ) 

if 

tangent thetatanθequals=eight fifteenths815​, 

piπless than<thetaθless than<StartFraction 3 pi Over 2 EndFraction3π2

Question content area bottom

Part 1

A.negative StartFraction 161 Over 289 EndFraction

negative StartFraction 161 Over 289 EndFraction−161289

B.StartFraction 161 Over 289 EndFraction

StartFraction 161 Over 289 EndFraction161289

C.StartFraction 240 Over 289 EndFraction

StartFraction 240 Over 289 EndFraction240289

D.negative StartFraction 240 Over 289 EndFraction

negative StartFraction 240 Over 289 EndFraction

To find \(\sin(2\theta)\) given that \(\tan(\theta) = \frac{8}{15}\) and \(\pi < \theta < \frac{3\pi}{2}\), we can use the double angle identity for sine:

\[
\sin(2\theta) = 2\sin(\theta)\cos(\theta)
\]

First, we need to find \(\sin(\theta)\) and \(\cos(\theta)\) using the given information.

### Step 1: Use \(\tan(\theta) = \frac{8}{15}\)

Recall that:

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

Since \(\tan(\theta) = \frac{8}{15}\), we have:

\[
\frac{\sin(\theta)}{\cos(\theta)} = \frac{8}{15}
\]

Let \(\sin(\theta) = 8k\) and \(\cos(\theta) = 15k\) for some constant \(k\).

### Step 2: Use the Pythagorean identity

The Pythagorean identity states that:

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

Substitute \(\sin(\theta) = 8k\) and \(\cos(\theta) = 15k\) into this identity:

\[
(8k)^2 + (15k)^2 = 1
\]

\[
64k^2 + 225k^2 = 1
\]

\[
289k^2 = 1
\]

\[
k^2 = \frac{1}{289}
\]

\[
k = \frac{1}{17}
\]

### Step 3: Find \(\sin(\theta)\) and \(\cos(\theta)\)

Now that we know \(k = \frac{1}{17}\), we can find:

\[
\sin(\theta) = 8k = \frac{8}{17}
\]

\[
\cos(\theta) = 15k = \frac{15}{17}
\]

### Step 4: Determine the signs of \(\sin(\theta)\) and \(\cos(\theta)\)

Since \(\pi < \theta < \frac{3\pi}{2}\), we know that \(\theta\) is in the third quadrant. In this quadrant, both sine and cosine are negative. Thus:

\[
\sin(\theta) = -\frac{8}{17}, \quad \cos(\theta) = -\frac{15}{17}
\]

### Step 5: Use the double angle identity

Now we can calculate \(\sin(2\theta)\):

\[
\sin(2\theta) = 2\sin(\theta)\cos(\theta)
\]

\[
\sin(2\theta) = 2 \left(-\frac{8}{17}\right) \left(-\frac{15}{17}\right)
\]

\[
\sin(2\theta) = 2 \times \frac{120}{289}
\]

\[
\sin(2\theta) = \frac{240}{289}
\]

### Final Answer

Thus, the value of \(\sin(2\theta)\) is:

\[
\boxed{\frac{240}{289}}
\]

So, the correct choice is:

**C. \(\frac{240}{289}\)**

---

Would you like to delve deeper into the steps or have any questions?

Here are some related questions you may find interesting:
1. How do we derive the double angle formula for sine?
2. How do we use the Pythagorean identity to find \(\sin(\theta)\) and \(\cos(\theta)\)?
3. What are the properties of trigonometric functions in different quadrants?
4. How can we apply the same method to find \(\cos(2\theta)\)?
5. What happens if \(\theta\) is in the first quadrant instead?

Tip: When working with trigonometric identities, always remember the signs of functions in different quadrants to avoid errors.

Find sin(2θ) if tan(θ) = 8/15 and π < θ < 3π/2.

Learn how to find sin(2θ) given tan(θ) = 8/15 with detailed steps. This problem involves the use of the double angle identity for sine and the Pythagorean identity. Perfect for high school students studying trigonometry.

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