Question

\( \sin^4 \frac{\pi}{8} + \sin^4 \frac{3\pi}{8} + \sin^4 \frac{5\pi}{8} + \sin^4 \frac{7\pi}{8} = \)

• A. \( \frac{1}{4} \)
• B. \( \frac{3}{8} \)
• C. \( \frac{3}{2} \)
• D. \( \frac{3}{4} \)

Hint:

Use the properties of trigonometric functions for related angles, such as \( \sin(\pi - \theta) = \sin \theta \) and \( \sin(\frac{\pi}{2} - \theta) = \cos \theta \), to simplify the expression. Also, use the identity \( \sin^2 \theta + \cos^2 \theta = 1 \) and algebraic manipulations to evaluate the sum.

Answer 1

The Correct Option is C

We need to evaluate \( \sin^4 \frac{\pi}{8} + \sin^4 \frac{3\pi}{8} + \sin^4 \frac{5\pi}{8} + \sin^4 \frac{7\pi}{8} \).
We know that \( \sin(\pi - \theta) = \sin \theta \).
So, \( \sin \frac{5\pi}{8} = \sin(\pi - \frac{3\pi}{8}) = \sin \frac{3\pi}{8} \) and \( \sin \frac{7\pi}{8} = \sin(\pi - \frac{\pi}{8}) = \sin \frac{\pi}{8} \).
The expression becomes \( \sin^4 \frac{\pi}{8} + \sin^4 \frac{3\pi}{8} + \sin^4 \frac{3\pi}{8} + \sin^4 \frac{\pi}{8} = 2 \left( \sin^4 \frac{\pi}{8} + \sin^4 \frac{3\pi}{8} \right) \).
We also know that \( \sin(\frac{\pi}{2} - \theta) = \cos \theta \).
So, \( \sin \frac{3\pi}{8} = \sin(\frac{\pi}{2} - \frac{\pi}{8}) = \cos \frac{\pi}{8} \).
The expression further simplifies to \( 2 \left( \sin^4 \frac{\pi}{8} + \cos^4 \frac{\pi}{8} \right) \).
We know that \( a^4 + b^4 = (a^2 + b^2)^2 - 2a^2 b^2 \).
Let \( a = \sin \frac{\pi}{8} \) and \( b = \cos \frac{\pi}{8} \).
Then \( a^2 + b^2 = \sin^2 \frac{\pi}{8} + \cos^2 \frac{\pi}{8} = 1 \).
So, \( \sin^4 \frac{\pi}{8} + \cos^4 \frac{\pi}{8} = (1)^2 - 2 \sin^2 \frac{\pi}{8} \cos^2 \frac{\pi}{8} = 1 - \frac{1}{2} (2 \sin \frac{\pi}{8} \cos \frac{\pi}{8})^2 = 1 - \frac{1}{2} (\sin \frac{\pi}{4})^2 \).
\( \sin \frac{\pi}{4} = \frac{1}{\sqrt{2}} \), so \( (\sin \frac{\pi}{4})^2 = \frac{1}{2} \).
Therefore, \( \sin^4 \frac{\pi}{8} + \cos^4 \frac{\pi}{8} = 1 - \frac{1}{2} \left( \frac{1}{2} \right) = 1 - \frac{1}{4} = \frac{3}{4} \).
Finally, the original expression is \( 2 \left( \frac{3}{4} \right) = \frac{3}{2} \).