Given sin(𝜃𝜃) = 3/4 where 𝜋/2 ≤ 𝜃 ≤ *π*, determine the exact value of sin (2𝜃).  Given cos (2𝜃𝜃)= − 7/8, where 2𝜃 is an angle in standard position with a terminal arm in quadrant 3, determine the exact value of cos (𝜃𝜃) and sin (𝜃𝜃)

### Part 1: Finding \(\sin(2\theta)\)

Given:
\[
\sin(\theta) = \frac{3}{4}
\]
where \(\frac{\pi}{2} \leq \theta \leq \pi\).

We know the identity for \(\sin(2\theta)\):
\[
\sin(2\theta) = 2\sin(\theta)\cos(\theta)
\]

First, we need to find \(\cos(\theta)\).

### Finding \(\cos(\theta)\)

We use the Pythagorean identity:
\[
\sin^2(\theta) + \cos^2(\theta) = 1
\]

Substituting \(\sin(\theta) = \frac{3}{4}\):
\[
\left(\frac{3}{4}\right)^2 + \cos^2(\theta) = 1
\]
\[
\frac{9}{16} + \cos^2(\theta) = 1
\]
\[
\cos^2(\theta) = 1 - \frac{9}{16} = \frac{7}{16}
\]
\[
\cos(\theta) = \pm \frac{\sqrt{7}}{4}
\]

Since \(\theta\) is in the second quadrant (\(\frac{\pi}{2} \leq \theta \leq \pi\)), where cosine is negative:
\[
\cos(\theta) = -\frac{\sqrt{7}}{4}
\]

### Now, calculate \(\sin(2\theta)\):

\[
\sin(2\theta) = 2\sin(\theta)\cos(\theta) = 2 \times \frac{3}{4} \times \left(-\frac{\sqrt{7}}{4}\right) = -\frac{3\sqrt{7}}{8}
\]

### Part 2: Finding \(\cos(\theta)\) and \(\sin(\theta)\)

Given:
\[
\cos(2\theta) = -\frac{7}{8}
\]
and \(2\theta\) is in the third quadrant.

The identities for \(\cos(2\theta)\) are:
\[
\cos(2\theta) = 2\cos^2(\theta) - 1
\]
or
\[
\cos(2\theta) = 1 - 2\sin^2(\theta)
\]

#### Solving for \(\cos(\theta)\):

Using \(\cos(2\theta) = 2\cos^2(\theta) - 1\):
\[
-\frac{7}{8} = 2\cos^2(\theta) - 1
\]
\[
2\cos^2(\theta) = \frac{1}{8}
\]
\[
\cos^2(\theta) = \frac{1}{16}
\]
\[
\cos(\theta) = \pm \frac{1}{4}
\]

Since \(\theta\) is in the second quadrant, \(\cos(\theta)\) is negative:
\[
\cos(\theta) = -\frac{1}{4}
\]

#### Solving for \(\sin(\theta)\):

Using the identity:
\[
\sin^2(\theta) + \cos^2(\theta) = 1
\]
\[
\sin^2(\theta) + \left(-\frac{1}{4}\right)^2 = 1
\]
\[
\sin^2(\theta) + \frac{1}{16} = 1
\]
\[
\sin^2(\theta) = \frac{15}{16}
\]
\[
\sin(\theta) = \pm \frac{\sqrt{15}}{4}
\]

Since \(\theta\) is in the second quadrant, \(\sin(\theta)\) is positive:
\[
\sin(\theta) = \frac{\sqrt{15}}{4}
\]

### Final Answers:
- \(\sin(2\theta) = -\frac{3\sqrt{7}}{8}\)
- \(\cos(\theta) = -\frac{1}{4}\)
- \(\sin(\theta) = \frac{\sqrt{15}}{4}\)

---

Would you like more details on any of these steps, or do you have any further questions? 

Here are some related questions:
1. What is the significance of the Pythagorean identity?
2. Can you find \(\tan(2\theta)\) given \(\sin(2\theta)\) and \(\cos(2\theta)\)?
3. How does the quadrant affect the sign of trigonometric functions?
4. What are the double angle identities for cosine and sine?
5. How would the solution change if \(\theta\) were in the fourth quadrant?
6. Can you derive \(\sin(2\theta)\) from \(\sin(\theta)\) and \(\cos(\theta)\) in a different way?
7. What is the importance of the interval \(\frac{\pi}{2} \leq \theta \leq \pi\) in this problem?
8. How can we verify the calculated values using a unit circle?

**Tip**: When working with trigonometric identities, always pay attention to the quadrant to determine the correct sign of the function values.

Explore how to find exact values of trigonometric functions sin(2𝜃), cos(𝜃), and sin(𝜃) using given conditions. Includes step-by-step solutions for angles in specific quadrants.

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