Question

If $\cos^{-1}x-\cos^{-1}\frac{y}{3}=\alpha$, where $-1\le x\le1$, $-3\le y\le3$, $x\le\frac{y}{3}$, 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. $\frac{4}{9}\sin^{2}\alpha$

Hint:

Logic Tip: You can often solve these generalized algebraic identity problems by plugging in arbitrary valid numbers! Let $x = 0$ and $y = 0$. Then $\cos^{-1}0 - \cos^{-1}0 = \alpha \implies \pi/2 - \pi/2 = 0 \implies \alpha = 0$. Now substitute these into the expression: $9(0)^2 - 0 + 0^2 = 0$. Check options for $\alpha=0$: $9\sin^2(0) = 0$. It matches perfectly.

Answer 1

The Correct Option is C

Concept:
Use the standard inverse trigonometric formula for the difference of two inverse cosines: $$\cos^{-1}A - \cos^{-1}B = \cos^{-1}\left(AB + \sqrt{1-A^2}\sqrt{1-B^2}\right)$$ Apply this formula to eliminate the inverse trigonometric functions, then manipulate the resulting algebraic equation by squaring both sides to isolate the desired expression.
Step 1: Apply the inverse cosine subtraction formula.
Given: $$\cos^{-1}x - \cos^{-1}\left(\frac{y}{3}\right) = \alpha$$ Using the formula where $A = x$ and $B = \frac{y}{3}$: $$\cos^{-1}\left(x\left(\frac{y}{3}\right) + \sqrt{1-x^2}\sqrt{1-\left(\frac{y}{3}\right)^2}\right) = \alpha$$ Take the cosine of both sides: $$\frac{xy}{3} + \sqrt{1-x^2}\sqrt{1-\frac{y^2}{9 = \cos\alpha$$
Step 2: Isolate the radical and square both sides.
To remove the square roots, isolate the radical term and simplify the fraction inside the root: $$\frac{xy}{3} + \frac{\sqrt{1-x^2}\sqrt{9-y^2{3} = \cos\alpha$$ Multiply the entire equation by 3: $$xy + \sqrt{1-x^2}\sqrt{9-y^2} = 3\cos\alpha$$ Isolate the radical: $$\sqrt{1-x^2}\sqrt{9-y^2} = 3\cos\alpha - xy$$ Square both sides: $$(1-x^2)(9-y^2) = (3\cos\alpha - xy)^2$$
Step 3: Expand and simplify to find the required expression.
Expand both sides: $$9 - y^2 - 9x^2 + x^2y^2 = 9\cos^2\alpha - 6xy\cos\alpha + x^2y^2$$ The $x^2y^2$ terms cancel out on both sides: $$9 - y^2 - 9x^2 = 9\cos^2\alpha - 6xy\cos\alpha$$ Rearrange the terms to group the variables on one side: $$9x^2 - 6xy\cos\alpha + y^2 = 9 - 9\cos^2\alpha$$ Factor out the 9 on the right side: $$9x^2 - 6xy\cos\alpha + y^2 = 9(1 - \cos^2\alpha)$$ Using the identity $1 - \cos^2\alpha = \sin^2\alpha$: $$9x^2 - 6xy\cos\alpha + y^2 = 9\sin^2\alpha$$