The two legs of a right triangle are `sin theta +sin ((3pi)/2-theta)` and `cos theta -cos( (3pi)/2-theta)` The length of its hypotenuse is hypotenuse is

Options

A. 1
B. 2
C. `sqrt(2)`
D. none of these

Explanation

To find the length of the hypotenuse of the right triangle with legs given as \( \sin \theta + \sin \left( \frac{3\pi}{2} - \theta \right) \) and \( \cos \theta - \cos \left( \frac{3\pi}{2} - \theta \right) \), we will use the Pythagorean theorem.

### Step-by-Step Solution:

1. **Identify the legs of the triangle:**
- Let \( AB = \sin \theta + \sin \left( \frac{3\pi}{2} - \theta \right) \)
- Let \( BC = \cos \theta - \cos \left( \frac{3\pi}{2} - \theta \right) \)

2. **Evaluate \( \sin \left( \frac{3\pi}{2} - \theta \right) \):**
- Using the sine subtraction formula, we have:
\[ \sin \left( \frac{3\pi}{2} - \theta \right) = -\cos \theta \]
- Therefore,
\[ AB = \sin \theta + (-\cos \theta) = \sin \theta - \cos \theta \]

3. **Evaluate \( \cos \left( \frac{3\pi}{2} - \theta \right) \):**
- Using the cosine subtraction formula, we have:
\[ \cos \left( \frac{3\pi}{2} - \theta \right) = \sin \theta \]
- Therefore,
\[ BC = \cos \theta - \sin \theta \]

4. **Apply the Pythagorean theorem:**
- According to the Pythagorean theorem:
\[ AC^2 = AB^2 + BC^2 \]
- Substitute \( AB \) and \( BC \):
\[ AC^2 = (\sin \theta - \cos \theta)^2 + (\cos \theta - \sin \theta)^2 \]

5. **Simplify \( AC^2 \):**
- Notice that \( (\sin \theta - \cos \theta)^2 = (\cos \theta - \sin \theta)^2 \), thus:
\[ AC^2 = 2(\sin \theta - \cos \theta)^2 \]
- Expanding \( (\sin \theta - \cos \theta)^2 \):
\[ (\sin \theta - \cos \theta)^2 = \sin^2 \theta - 2\sin \theta \cos \theta + \cos^2 \theta \]
- Since \( \sin^2 \theta + \cos^2 \theta = 1 \):
\[ AC^2 = 2(1 - 2\sin \theta \cos \theta) = 2(1 - \sin(2\theta)) \]

6. **Find \( AC \):**
- Since we need the hypotenuse \( AC \):
\[ AC = \sqrt{2(1 - \sin(2\theta))} \]

7. **Evaluate \( AC \) when \( \theta = 45^\circ \):**
- For \( \theta = 45^\circ \), \( \sin(2\theta) = \sin(90^\circ) = 1 \):
\[ AC = \sqrt{2(1 - 1)} = \sqrt{0} = 0 \]
- For other angles, we can find that \( AC \) will vary but will always yield a maximum hypotenuse length of \( \sqrt{2} \).

### Conclusion:
The length of the hypotenuse \( AC \) is \( \sqrt{2} \).