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Answer the following.
If cos⁻¹x/2 + cos⁻¹y/3 = θ, then prove that 9x² - 12xycosθ + 4y² = 36sin²θ.
If $\cos^{-1}\frac{x}{2} + \cos^{-1}\frac{y}{3} = \theta$, then prove that $9x^2 - 12xy\cos\theta + 4y^2 = 36\sin^2\theta$.
Given: $\cos^{-1}\frac{x}{2} + \cos^{-1}\frac{y}{3} = \theta$
To show: $9x^2 - 12xy\cos\theta + 4y^2 = 36\sin^2\theta$
Answer
Answer
Hence $9x^2 - 12xy\cos\theta + 4y^2 = 36\sin^2\theta$.
- Let $\cos^{-1}\frac{x}{2} = \alpha$ and $\cos^{-1}\frac{y}{3} = \beta$, so $\cos\alpha = \frac{x}{2}$, $\cos\beta = \frac{y}{3}$ and $\alpha + \beta = \theta$.
- $\sin\alpha = \sqrt{1 - \frac{x^2}{4}}$ and $\sin\beta = \sqrt{1 - \frac{y^2}{9}}$, by $\sin^2 A = 1 - \cos^2 A$.
- $\cos\theta = \cos(\alpha + \beta) = \cos\alpha\cos\beta - \sin\alpha\sin\beta$, by the addition formula.
- $\cos\theta = \frac{xy}{6} - \sqrt{\frac{(4 - x^2)(9 - y^2)}{36}}$.
- Rearrange: $\frac{xy}{6} - \cos\theta = \frac{\sqrt{(4 - x^2)(9 - y^2)}}{6}$.
- Squaring both sides: $\frac{x^2y^2}{36} - \frac{xy\cos\theta}{3} + \cos^2\theta = \frac{(4 - x^2)(9 - y^2)}{36}$.
- Multiply by $36$: $x^2y^2 - 12xy\cos\theta + 36\cos^2\theta = 36 - 4y^2 - 9x^2 + x^2y^2$.
- Cancel $x^2y^2$ and collect: $9x^2 - 12xy\cos\theta + 4y^2 = 36 - 36\cos^2\theta$.
- Using $1 - \cos^2\theta = \sin^2\theta$: $9x^2 - 12xy\cos\theta + 4y^2 = 36\sin^2\theta$.