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

We are given: \[ y = \cos \left( \cos^{-1} \frac{1}{2} + \cos^{-1} \frac{x}{2} \right) \] Since \( \cos^{-1} \frac{1}{2} = \frac{\pi}{3} \), we get: \[ y = \cos \left( \frac{\pi}{3} + \cos^{-1} \frac{x}{2} \right) \] Let \( \theta = \cos^{-1} \frac{x}{2} \), so we have: \[ y = \cos \left( \frac{\pi}{3} + \theta \right) \] Using the addition formula for cosine: \[ y = \frac{1}{2} \cos \theta - \sqrt{3} \sin \theta \] Now, squaring both sides: \[ y^2 = \left( \frac{1}{2} \cos \theta - \sqrt{3} \sin \theta \right)^2 \] Expanding the squares: \[ y^2 = \frac{1}{4} \cos^2 \theta + 3 \sin^2 \theta - \sqrt{3} \sin 2\theta \] Now, use the identity \( \sin^2 \theta = 1 - \cos^2 \theta \) to further simplify: \[ y^2 = \frac{1}{4} \cos^2 \theta + 3 (1 - \cos^2 \theta) - \sqrt{3} \sin 2\theta \] \[ y^2 = \frac{1}{4} \cos^2 \theta + 3 - 3 \cos^2 \theta - \sqrt{3} \sin 2\theta \] \[ y^2 = 3 - \frac{11}{4} \cos^2 \theta - \sqrt{3} \sin 2\theta \] Next, for \( (x - y)^2 + 3y^2 \): \[ (x - y)^2 + 3y^2 = 3 \]

Answer 2

We are given \( y = \cos\!\left(\frac{\pi}{3} + \cos^{-1}\!\frac{x}{2}\right) \) and asked to evaluate the expression \( (x - y)^2 + 3y^2 \) in terms of \(x\).

Concept Used:

Use the cosine addition identity and the relation between sine and cosine via inverse cosine. For any angle \( \theta \):

\[ \cos(\alpha+\beta)=\cos\alpha\cos\beta-\sin\alpha\sin\beta,\quad \text{and if } \cos\theta = u,\ \text{then } \sin\theta=\sqrt{1-u^2}\ (\theta\in[0,\pi]). \]

Step-by-Step Solution:

Step 1: Let \( \theta = \cos^{-1}\!\left(\frac{x}{2}\right) \). Then \( \cos\theta = \frac{x}{2} \) and, since \( \theta \in [0,\pi] \),

\[ \sin\theta = \sqrt{1-\cos^2\theta} = \sqrt{1-\left(\frac{x}{2}\right)^2} = \frac{\sqrt{4-x^2}}{2}. \]

Step 2: Expand \( y = \cos\!\left(\frac{\pi}{3}+\theta\right) \) using \( \cos\frac{\pi}{3}=\frac12 \) and \( \sin\frac{\pi}{3}=\frac{\sqrt{3}}{2} \):

\[ y=\cos\frac{\pi}{3}\cos\theta - \sin\frac{\pi}{3}\sin\theta = \frac12\cdot\frac{x}{2} - \frac{\sqrt{3}}{2}\cdot\frac{\sqrt{4-x^2}}{2} = \frac{x}{4} - \frac{\sqrt{3}}{4}\sqrt{4-x^2}. \]

Step 3: Set \( a=\frac{x}{4} \) and \( b=\frac{\sqrt{3}}{4}\sqrt{4-x^2} \). Then \( y=a-b \) and

\[ x-y = x-(a-b) = \frac{3x}{4} + b. \]

Step 4: Compute the required expression:

\[ (x-y)^2 + 3y^2 = \left(\frac{3x}{4}+b\right)^2 + 3(a-b)^2. \]

Expand and group like terms:

\[ \left(\frac{3x}{4}\right)^2 + \frac{3x}{2}\,b + b^2 \;+\; 3a^2 - 6ab + 3b^2 = \frac{9x^2}{16} + 3a^2 + 4b^2 + \left(\frac{3x}{2}b - 6ab\right). \]

Step 5: Note the mixed terms cancel since \( a=\tfrac{x}{4} \):

\[ \frac{3x}{2}b - 6ab = \frac{3x}{2}b - 6\left(\frac{x}{4}\right)b = 0. \]

Also compute \( 3a^2 = 3\left(\frac{x}{4}\right)^2 = \frac{3x^2}{16} \) and \( 4b^2 = 4\left(\frac{3}{16}(4-x^2)\right) = \frac{12}{16}(4-x^2)=3-\frac{3x^2}{4} \). Hence,

\[ (x-y)^2 + 3y^2 = \left(\frac{9x^2}{16} + \frac{3x^2}{16}\right) + \left(3 - \frac{3x^2}{4}\right) = \frac{12x^2}{16} + 3 - \frac{3x^2}{4} = \frac{3x^2}{4} + 3 - \frac{3x^2}{4} = 3. \]

Final Computation & Result

The expression simplifies to the constant value 3, independent of \(x\):

\[ \boxed{(x - y)^2 + 3y^2 = 3}. \]