Question

If $ y = \cos \left( \frac{\pi}{3} + \cos^{-1} \frac{x}{2} \right) $, then $ (x - y)^2 + 3y^2 $ is equal to ________.

Hint:

In such trigonometric problems, use trigonometric identities to simplify expressions and solve for the desired quantity. Ensure that all terms are properly expanded and simplified.

Answer 1

Correct Answer: 3

The given expression is: \[ y = \cos \left( \cos^{-1} \frac{1}{2} + \cos^{-1} \frac{x}{2} \right) \] Given that \( \cos^{-1} \frac{1}{2} = \frac{\pi}{3} \), the expression becomes: \[ y = \cos \left( \frac{\pi}{3} + \cos^{-1} \frac{x}{2} \right) \] Let \( \theta = \cos^{-1} \frac{x}{2} \). Then: \[ y = \cos \left( \frac{\pi}{3} + \theta \right) \] Applying the cosine addition formula: \[ y = \frac{1}{2} \cos \theta - \sqrt{3} \sin \theta \] Squaring both sides: \[ y^2 = \left( \frac{1}{2} \cos \theta - \sqrt{3} \sin \theta \right)^2 \] Expanding the squared term: \[ y^2 = \frac{1}{4} \cos^2 \theta + 3 \sin^2 \theta - \sqrt{3} \sin 2\theta \] Using the identity \( \sin^2 \theta = 1 - \cos^2 \theta \): \[ y^2 = \frac{1}{4} \cos^2 \theta + 3 (1 - \cos^2 \theta) - \sqrt{3} \sin 2\theta \] \[ y^2 = \frac{1}{4} \cos^2 \theta + 3 - 3 \cos^2 \theta - \sqrt{3} \sin 2\theta \] \[ y^2 = 3 - \frac{11}{4} \cos^2 \theta - \sqrt{3} \sin 2\theta \] For the expression \( (x - y)^2 + 3y^2 \): \[ (x - y)^2 + 3y^2 = 3 \]