Question

If \( \theta \) is an acute angle and \( 2\sin^2 \theta = \cos^4 \frac{\pi}{8} + \sin^4 \frac{3\pi}{8} + \cos^4 \frac{5\pi}{8} + \sin^4 \frac{7\pi}{8} \), then \( \theta \) is:

• A. \(\frac{\pi}{6}\)
• B. \(\frac{\pi}{4}\)
• C. \(\frac{\pi}{3}\)
• D. \(\frac{\pi}{8}\)

Hint:

Always verify trigonometric identities and simplify based on known angle values and their symmetries. Double-check calculations especially when equating complex trigonometric expressions.

Answer 1

The Correct Option is C

To evaluate the right-hand side, we start by recognizing the symmetry and identity relations between the given angles: \[ \cos \frac{\pi}{8} = \sin \frac{7\pi}{8}, \quad \sin \frac{3\pi}{8} = \cos \frac{5\pi}{8} \] Thus, we can simplify the expression by noting these values are equal due to their complementary nature and periodic properties of sine and cosine functions: \[ \cos^4 \frac{\pi}{8} + \sin^4 \frac{3\pi}{8} + \cos^4 \frac{5\pi}{8} + \sin^4 \frac{7\pi}{8} = 2 \cos^4 \frac{\pi}{8} + 2 \sin^4 \frac{3\pi}{8} \] Using the identity \(x^4 + y^4 = (x^2 + y^2)^2 - 2x^2y^2\) for \(x = \cos \frac{\pi}{8}\) and \(y = \sin \frac{3\pi}{8}\): \[ 2 \left(\cos^4 \frac{\pi}{8} + \sin^4 \frac{3\pi}{8}\right) = 2 \left( (\cos^2 \frac{\pi}{8} + \sin^2 \frac{3\pi}{8})^2 - 2 \cos^2 \frac{\pi}{8} \sin^2 \frac{3\pi}{8} \right) \] Since \( \cos^2 \frac{\pi}{8} + \sin^2 \frac{3\pi}{8} = 1 \) and \[ \cos^2 \frac{\pi}{8} \sin^2 \frac{3\pi}{8} = \left( \frac{\sqrt{2 - \sqrt{2}}}{2} \right)^2 \left( \frac{\sqrt{2 + \sqrt{2}}}{2} \right)^2 = \frac{1}{4} \] \[ 2 \left( 1^2 - 2 \times \frac{1}{4} \right) = 2 \left( 1 - \frac{1}{2} \right) = 1 \] Given the equation is \(2\sin^2 \theta = 1\), solve for \(\theta\): \[ \sin^2 \theta = \frac{1}{2} \Rightarrow \sin \theta = \frac{\sqrt{2}}{2} \] Since \(\theta\) is acute and \(\sin \frac{\pi}{4} = \frac{\sqrt{2}}{2}\), the solution for \(\theta\) is \(\frac{\pi}{4}\), not \(\frac{\pi}{3}\) as initially suggested.