Problem
If \[ \cos^{-1}\left(\frac{x}{2}\right) + \cos^{-1}\left(\frac{y}{3}\right) = \alpha \] prove that \[ 9x^2 – 12xy\cos\alpha + 4y^2 = 36\sin^2\alpha \]
Solution
Let:
\[ A = \cos^{-1}\left(\frac{x}{2}\right), \quad B = \cos^{-1}\left(\frac{y}{3}\right) \]
\[ A + B = \alpha \]
Step 1: Express cos A, cos B
\[ \cos A = \frac{x}{2}, \quad \cos B = \frac{y}{3} \]
Step 2: Use cos(A+B)
\[ \cos\alpha = \cos A \cos B – \sin A \sin B \]
\[ = \frac{x}{2}\cdot\frac{y}{3} – \sqrt{1-\frac{x^2}{4}}\cdot\sqrt{1-\frac{y^2}{9}} \]
Step 3: Rearrange
\[ \sqrt{1-\frac{x^2}{4}}\cdot\sqrt{1-\frac{y^2}{9}} = \frac{xy}{6} – \cos\alpha \]
Step 4: Square both sides
\[ \left(1-\frac{x^2}{4}\right)\left(1-\frac{y^2}{9}\right) = \left(\frac{xy}{6} – \cos\alpha\right)^2 \]
Step 5: Expand LHS
\[ 1 – \frac{x^2}{4} – \frac{y^2}{9} + \frac{x^2y^2}{36} \]
Step 6: Expand RHS
\[ \frac{x^2y^2}{36} – \frac{xy}{3}\cos\alpha + \cos^2\alpha \]
Step 7: Cancel common terms
\[ 1 – \frac{x^2}{4} – \frac{y^2}{9} = – \frac{xy}{3}\cos\alpha + \cos^2\alpha \]
Step 8: Multiply by 36
\[ 36 – 9x^2 – 4y^2 = -12xy\cos\alpha + 36\cos^2\alpha \]
Step 9: Rearrange
\[ 9x^2 – 12xy\cos\alpha + 4y^2 = 36(1 – \cos^2\alpha) \]
\[ = 36\sin^2\alpha \]
Final Result
\[ \boxed{9x^2 – 12xy\cos\alpha + 4y^2 = 36\sin^2\alpha} \]
Explanation
Use cos(A+B), square both sides, expand, and simplify carefully.