Question

If \( \cos^{-1}\frac{x}{2} + \cos^{-1}\frac{y}{3} = \phi \), then \( 9x^2 - 12xy\cos\phi + 4y^2 \) is equal to

• A. \( -36\sin^2\phi \)
• B. \( 36\sin^2\phi \)
• C. \( 36\cos^2\phi \)
• D. 36

Hint:

This is a standard identity derived from the \( \cos(A+B) \) formula. For an equation of the form \( \cos^{-1}x + \cos^{-1}y = \theta \), the resulting algebraic relation is always \( x^2 - 2xy\cos\theta + y^2 = \sin^2\theta \). You can adapt this general form to the given problem by setting \( x' = x/2 \) and \( y' = y/3 \).

Answer 1

The Correct Option is B

Let \( A = \cos^{-1}\frac{x}{2} \) and \( B = \cos^{-1}\frac{y}{3} \). This means \( \cos A = \frac{x}{2} \) and \( \cos B = \frac{y}{3} \). We can find \( \sin A \) and \( \sin B \) using \( \sin\theta = \sqrt{1-\cos^2\theta} \). \[ \sin A = \sqrt{1 - \frac{x^2}{4}} = \frac{\sqrt{4-x^2}}{2} \] \[ \sin B = \sqrt{1 - \frac{y^2}{9}} = \frac{\sqrt{9-y^2}}{3} \] We are given \( A+B = \phi \). Apply cosine to both sides: \( \cos(A+B) = \cos\phi \). Use the cosine addition formula: \( \cos A \cos B - \sin A \sin B = \cos\phi \). Substitute the expressions for the trig functions: \[ \left(\frac{x}{2}\right)\left(\frac{y}{3}\right) - \left(\frac{\sqrt{4-x^2}}{2}\right)\left(\frac{\sqrt{9-y^2}}{3}\right) = \cos\phi \] \[ \frac{xy}{6} - \frac{\sqrt{(4-x^2)(9-y^2)}}{6} = \cos\phi \] \[ xy - \sqrt{(4-x^2)(9-y^2)} = 6\cos\phi \] \[ xy - 6\cos\phi = \sqrt{(4-x^2)(9-y^2)} \] Square both sides to eliminate the square root: \[ (xy - 6\cos\phi)^2 = (4-x^2)(9-y^2) \] \[ x^2y^2 - 12xy\cos\phi + 36\cos^2\phi = 36 - 4y^2 - 9x^2 + x^2y^2 \] Cancel \( x^2y^2 \) from both sides and rearrange: \[ -12xy\cos\phi + 36\cos^2\phi = 36 - 4y^2 - 9x^2 \] Move all the \( x, y \) terms to the left side: \[ 9x^2 - 12xy\cos\phi + 4y^2 = 36 - 36\cos^2\phi \] \[ 9x^2 - 12xy\cos\phi + 4y^2 = 36(1 - \cos^2\phi) \] Using the identity \( \sin^2\phi + \cos^2\phi = 1 \), we get: \[ 9x^2 - 12xy\cos\phi + 4y^2 = 36\sin^2\phi \]