Question
If
\[ \cos^{-1}\left(\frac{x}{3}\right) + \cos^{-1}\left(\frac{y}{2}\right) = \frac{\theta}{2} \]
Find:
\[ 4x^2 – 12xy\cos\frac{\theta}{2} + 9y^2 \]
Solution
Let
\[ \cos^{-1}\left(\frac{x}{3}\right) = A,\quad \cos^{-1}\left(\frac{y}{2}\right) = B \]
Then,
\[ A + B = \frac{\theta}{2} \]
So,
\[ \cos A = \frac{x}{3}, \quad \cos B = \frac{y}{2} \]
We use identity:
\[ \cos(A + B) = \cos A \cos B – \sin A \sin B \]
\[ \cos\frac{\theta}{2} = \frac{x}{3} \cdot \frac{y}{2} – \sqrt{1 – \frac{x^2}{9}} \cdot \sqrt{1 – \frac{y^2}{4}} \]
Using standard symmetric identity result:
\[ \frac{x^2}{9} – \frac{2xy}{6}\cos\frac{\theta}{2} + \frac{y^2}{4} = \sin^2\frac{\theta}{2} \]
Multiply both sides by 36:
\[ 4x^2 – 12xy\cos\frac{\theta}{2} + 9y^2 = 36\sin^2\frac{\theta}{2} \]
Final Answer:
\[ \boxed{36\sin^2\frac{\theta}{2}} \]
Key Concept
Use cosine addition identity and convert into symmetric algebraic form.