Question

If $\cos^{-1} x - \cos^{-1} \frac{y}{3} = \alpha$, where $-1 \le x \le 1$, $-3 \le y \le 3$, $x \le \frac{y}{3}$, then for all $x, y$, $9x^2 - 6xy \cos \alpha + y^2$ is equal to}

• A. $\sin^2 \alpha$
• B. $3 \sin^2 \alpha$
• C. $9 \sin^2 \alpha$
• D. $\frac{1}{9} \sin^2 \alpha$

Hint:

Use the formula for the difference of inverse cosines. Be careful with algebraic manipulation and squaring both sides to eliminate square roots. Remember the identity $\sin^2 \theta + \cos^2 \theta = 1$.

Answer 1

The Correct Option is C

Step 1: Apply the difference formula for inverse cosines. Given the equation: \[ \cos^{-1} x - \cos^{-1} \frac{y}{3} = \alpha \] Using the formula $\cos^{-1} A - \cos^{-1} B = \cos^{-1}(AB + \sqrt{1-A^2}\sqrt{1-B^2})$: \[ \cos^{-1}\left(x \cdot \frac{y}{3} + \sqrt{1-x^2}\sqrt{1-\left(\frac{y}{3}\right)^2}\right) = \alpha \] \[ x \frac{y}{3} + \sqrt{1-x^2}\sqrt{1-\frac{y^2}{9 = \cos \alpha \]
Step 2: Isolate the square root terms and square both sides. Move the term $\frac{xy}{3}$ to the right side: \[ \sqrt{1-x^2}\sqrt{1-\frac{y^2}{9 = \cos \alpha - \frac{xy}{3} \] Square both sides to eliminate the square roots: \[ \left(\sqrt{1-x^2}\sqrt{1-\frac{y^2}{9\right)^2 = \left(\cos \alpha - \frac{xy}{3}\right)^2 \] \[ (1-x^2)\left(1-\frac{y^2}{9}\right) = \cos^2 \alpha - 2\left(\frac{xy}{3}\right)\cos \alpha + \left(\frac{xy}{3}\right)^2 \]
Step 3: Expand and simplify the equation. \[ 1 - \frac{y^2}{9} - x^2 + \frac{x^2y^2}{9} = \cos^2 \alpha - \frac{2xy}{3}\cos \alpha + \frac{x^2y^2}{9} \] Notice that the term $\frac{x^2y^2}{9}$ appears on both sides, so it can be cancelled out: \[ 1 - \frac{y^2}{9} - x^2 = \cos^2 \alpha - \frac{2xy}{3}\cos \alpha \]
Step 4: Rearrange the terms to match the required expression. Multiply the entire equation by 9 to clear the denominators: \[ 9(1) - 9\left(\frac{y^2}{9}\right) - 9(x^2) = 9(\cos^2 \alpha) - 9\left(\frac{2xy}{3}\cos \alpha\right) \] \[ 9 - y^2 - 9x^2 = 9\cos^2 \alpha - 6xy\cos \alpha \] Rearrange the terms to obtain the expression $9x^2 - 6xy \cos \alpha + y^2$: \[ 9x^2 - 6xy \cos \alpha + y^2 = 9 - 9\cos^2 \alpha \]
Step 5: Use trigonometric identities to find the final value. Factor out 9 from the right side: \[ 9x^2 - 6xy \cos \alpha + y^2 = 9(1 - \cos^2 \alpha) \] Using the identity $\sin^2 \alpha + \cos^2 \alpha = 1$, which implies $1 - \cos^2 \alpha = \sin^2 \alpha$: \[ 9x^2 - 6xy \cos \alpha + y^2 = 9\sin^2 \alpha \]