`sin^(4) ""(pi)/(8)+ sin^(4) ""(3pi)/(8) + sin^(4) ""(5 pi)/(8) + sin^(4) ""(7 pi)/(8)` is equal to

Options

A. 1
B. `(3)/(2)`
C. 2
D. `(1)/(4)`

Answer

To solve the expression \( \sin^4 \left( \frac{\pi}{8} \right) + \sin^4 \left( \frac{3\pi}{8} \right) + \sin^4 \left( \frac{5\pi}{8} \right) + \sin^4 \left( \frac{7\pi}{8} \right) \), we will use some trigonometric identities and properties.

### Step-by-Step Solution:

1. **Rewrite the Sine Terms**:
We can use the identity \( \sin(\pi - x) = \sin(x) \) to simplify the terms:
\[ \sin\left( \frac{5\pi}{8} \right) = \sin\left( \pi - \frac{3\pi}{8} \right) = \sin\left( \frac{3\pi}{8} \right) \]
\[ \sin\left( \frac{7\pi}{8} \right) = \sin\left( \pi - \frac{\pi}{8} \right) = \sin\left( \frac{\pi}{8} \right) \]

Therefore, we can rewrite the expression as:
\[ \sin^4 \left( \frac{\pi}{8} \right) + \sin^4 \left( \frac{3\pi}{8} \right) + \sin^4 \left( \frac{3\pi}{8} \right) + \sin^4 \left( \frac{\pi}{8} \right) \]
Which simplifies to:
\[ 2\sin^4 \left( \frac{\pi}{8} \right) + 2\sin^4 \left( \frac{3\pi}{8} \right) \]
Thus, we can factor out the 2:
\[ 2 \left( \sin^4 \left( \frac{\pi}{8} \right) + \sin^4 \left( \frac{3\pi}{8} \right) \right) \]

2. **Use the Identity for Sine Squared**:
We can use the identity \( \sin^2 x + \cos^2 x = 1 \) to express \( \sin^4 x \):
\[ \sin^4 x = (\sin^2 x)^2 = \left( \sin^2 x \right)^2 = \left( 1 - \cos^2 x \right)^2 \]
Therefore, we can express \( \sin^4 \left( \frac{\pi}{8} \right) \) and \( \sin^4 \left( \frac{3\pi}{8} \right) \) as:
\[ \sin^4 \left( \frac{\pi}{8} \right) = \left( \sin^2 \left( \frac{\pi}{8} \right) \right)^2 \]
\[ \sin^4 \left( \frac{3\pi}{8} \right) = \left( \sin^2 \left( \frac{3\pi}{8} \right) \right)^2 \]

3. **Using the Double Angle Formula**:
We can use the double angle formula:
\[ \sin(2x) = 2\sin(x)\cos(x) \]
Thus,
\[ \sin^2 \left( \frac{3\pi}{8} \right) = \cos^2 \left( \frac{\pi}{8} \right) \]

4. **Combine the Terms**:
Now, we can combine the squares:
\[ \sin^4 \left( \frac{\pi}{8} \right) + \sin^4 \left( \frac{3\pi}{8} \right) = \sin^4 \left( \frac{\pi}{8} \right) + \cos^4 \left( \frac{\pi}{8} \right) \]
Using the identity \( a^4 + b^4 = (a^2 + b^2)^2 - 2a^2b^2 \):
\[ = \left( \sin^2 \left( \frac{\pi}{8} \right) + \cos^2 \left( \frac{\pi}{8} \right) \right)^2 - 2\sin^2 \left( \frac{\pi}{8} \right) \cos^2 \left( \frac{\pi}{8} \right) \]
Since \( \sin^2 \left( \frac{\pi}{8} \right) + \cos^2 \left( \frac{\pi}{8} \right) = 1 \):
\[ = 1 - 2\sin^2 \left( \frac{\pi}{8} \right) \cos^2 \left( \frac{\pi}{8} \right) \]

5. **Final Calculation**:
We know that \( \sin(2x) = 2\sin(x)\cos(x) \):
\[ = 1 - \frac{1}{2} \sin^2 \left( \frac{\pi}{4} \right) = 1 - \frac{1}{2} \cdot \frac{1}{2} = 1 - \frac{1}{4} = \frac{3}{4} \]
Therefore, multiplying by 2:
\[ 2 \left( \frac{3}{4} \right) = \frac{3}{2} \]

### Final Answer:
\[ \sin^4 \left( \frac{\pi}{8} \right) + \sin^4 \left( \frac{3\pi}{8} \right) + \sin^4 \left( \frac{5\pi}{8} \right) + \sin^4 \left( \frac{7\pi}{8} \right) = \frac{3}{2} \]