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 [2023]

• A. $\sin^2\alpha$
• B. $3\sin^2\alpha$
• C. $9\sin^2\alpha$
• D. $\frac{4}{9}\sin^2\alpha$

Answer 1

Answer: (C)

$9\sin^2\alpha

Answer 2

Let

$A = \cos^{-1}(x) \quad \text{and} \quad B = \cos^{-1}\left(\frac{y}{3}\right)$.

Then,

$A - B = \alpha \quad \Rightarrow \quad A = \alpha + B$.

So,

$x = \cos A = \cos(\alpha+B) = \cos\alpha \cos B - \sin\alpha \sin B$.

Since $\cos B = \frac{y}{3}$, we have:

$x = \frac{y\cos\alpha}{3} - \sin\alpha \sin B$.

Multiply by 3:

$3x = y\cos\alpha - 3\sin\alpha \sin B$.

Now, the expression to evaluate is:

$9x^2 - 6xy\cos\alpha + y^2$.

Notice that

$(3x - y\cos\alpha)^2 = 9x^2 - 6xy\cos\alpha + y^2\cos^2\alpha$.

Thus,

$9x^2 - 6xy\cos\alpha + y^2 = (3x - y\cos\alpha)^2 + y^2\sin^2\alpha$.

Substitute for $3x - y\cos\alpha$:

$3x - y\cos\alpha = -3\sin\alpha \sin B \quad \Rightarrow \quad (3x - y\cos\alpha)^2 = 9\sin^2\alpha \sin^2B$.

Also, since $\cos B=\frac{y}{3}$, we have $y = 3\cos B$ and so:

$y^2\sin^2\alpha = 9\cos^2B\sin^2\alpha$.

Therefore,

$9x^2 - 6xy\cos\alpha + y^2 = 9\sin^2\alpha\sin^2B + 9\sin^2\alpha\cos^2B = 9\sin^2\alpha(\sin^2B+\cos^2B)$.

Since $\sin^2B+\cos^2B=1$, we obtain:

$9x^2 - 6xy\cos\alpha + y^2 = 9\sin^2\alpha$.