Given that the inverse trigonometric function assumes principal values only. Let x, y be any two rea

Q: Wait, why do we convert $\sin^{-1} y$ to $\cos^{-1} y$? Can't we just use the sine addition formula?

Great question! While you could technically convert everything to sine, the target expression $x^2 + y^2 + 2xy \sin \alpha$ is already structured in a way that aligns perfectly with the cosine addition formula. By converting to $\cos^{-1}$, we naturally generate the $xy$ and $\sqrt{1-x^2}\sqrt{1-y^2}$ terms that appear in the expansion of the target expression. It is all about choosing the path of least resistance!

Q: Why do we have to isolate the radical term before squaring? Can't I just square both sides immediately?

You *could*, but it would be a trap! If you square the equation $xy - \sqrt{1-x^2}\sqrt{1-y^2} = -\sin \alpha$ directly, you will end up with a complex cross-term involving the square root, which forces you to square a second time. By isolating the radical first, you ensure that the squaring process eliminates the roots in one clean step, leading directly to the simplified algebraic form.

Q: How do I know the minimum value is 0? Is it possible for the expression to be negative?

Excellent intuition. Look at the final form we derived: $\cos^2 \alpha$. Since $\cos^2 \alpha$ is a squared real number, it can never be negative. It is always greater than or equal to zero. Within the given domain of $\alpha \in [-\frac{\pi}{2}, \pi]$, the function $\cos \alpha$ hits zero at $\alpha = -\frac{\pi}{2}$ and $\alpha = \frac{\pi}{2}$. Thus, the minimum value is indeed 0.

Question

Given that the inverse trigonometric function assumes principal values only. Let $x, y$ be any two real numbers in $[-1, 1]$ such that $\cos^{-1} x - \sin^{-1} y = \alpha, -\frac{\pi}{2} \le \alpha \le \pi$. Then, the minimum value of $x^2 + y^2 + 2xy \sin \alpha$ is

Options:

(A) 0

(B) -1

(C) $1/2$

(D) $-1/2$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing at the base of a mountain, looking at a complex expression: $x^2 + y^2 + 2xy \sin \alpha$. It looks intimidating, but in the world of JEE Advanced, complexity is often just a mask for hidden elegance.

Our journey begins with the given equation:
$$\cos^{-1} x - \sin^{-1} y = \alpha$$

The first hurdle is the mix of inverse trigonometric functions. We cannot easily combine a $\cos^{-1}$ and a $\sin^{-1}$. However, we hold the key in our toolkit: the identity $\sin^{-1} y = \frac{\pi}{2} - \cos^{-1} y$.

By substituting this, we transform our equation into:
$$\cos^{-1} x - \left( \frac{\pi}{2} - \cos^{-1} y ight) = \alpha$$

With a quick rearrangement, we get:
$$\cos^{-1} x + \cos^{-1} y = \frac{\pi}{2} + \alpha$$

Now, the symmetry is beautiful. We have two inverse cosines on the left, ready to be combined.

### The Algebraic Dance

To merge these terms, we invoke the addition formula for inverse cosines:
$$\cos^{-1} A + \cos^{-1} B = \cos^{-1} \left( AB - \sqrt{1-A^2}\sqrt{1-B^2} ight)$$

Applying this to our variables $x$ and $y$, the equation becomes:
$$\cos^{-1} \left( xy - \sqrt{1-x^2}\sqrt{1-y^2} ight) = \frac{\pi}{2} + \alpha$$

To strip away the inverse cosine, we take the cosine of both sides. The left side simplifies to $xy - \sqrt{1-x^2}\sqrt{1-y^2}$. On the right, we have $\cos \left( \frac{\pi}{2} + \alpha ight)$.

Using the allied angle identity, $\cos \left( \frac{\pi}{2} + 	heta ight) = -\sin 	heta$, the right side becomes $-\sin \alpha$. We are left with:
$$xy - \sqrt{1-x^2}\sqrt{1-y^2} = -\sin \alpha$$

### The Radical Trap

Here is where many students stumble. We have square roots, and our instinct is to square everything immediately. But hold on! If you square now, you will create a mess of cross-terms.

The master move is to isolate the radical. Let's move $-\sin \alpha$ to the left and the radical to the right:
$$xy + \sin \alpha = \sqrt{1-x^2}\sqrt{1-y^2}$$

Now, when we square both sides, the radical vanishes cleanly:
$$(xy + \sin \alpha)^2 = (1-x^2)(1-y^2)$$

Expanding both sides, we get:
$$x^2y^2 + 2xy \sin \alpha + \sin^2 \alpha = 1 - x^2 - y^2 + x^2y^2$$

Notice the $x^2y^2$ on both sides? They cancel out perfectly, leaving us with:
$$2xy \sin \alpha + \sin^2 \alpha = 1 - x^2 - y^2$$

### The Elegant Collapse

We are almost at the summit. Rearrange the terms to isolate our target expression:
$$x^2 + y^2 + 2xy \sin \alpha = 1 - \sin^2 \alpha$$

By the fundamental identity $\sin^2 \alpha + \cos^2 \alpha = 1$, we know that $1 - \sin^2 \alpha = \cos^2 \alpha$. Our massive, terrifying expression has collapsed into the simple, elegant $\cos^2 \alpha$.

The problem is now reduced to finding the minimum value of $\cos^2 \alpha$ over the interval $[-\frac{\pi}{2}, \pi]$. Since $\cos^2 \alpha$ is always non-negative, its minimum is 0.

Does it reach 0 in our interval? Yes, at $\alpha = -\frac{\pi}{2}$ and $\alpha = \frac{\pi}{2}$. Thus, the **minimum value is 0**. You have navigated the complexity and found the truth hidden beneath.

Select Answer:

Visualized Solution

$Analyze the Given Equation$

$Convert sin^{- 1} y to cos^{- 1} y$

$Rearrange the Equation$

$Apply Addition Formula$

$Take Cosine on Both Sides$

$Isolate the Radical Term$

$Square Both Sides$

$Expand and Simplify$

$Isolate the Target Expression$

$Final Form: cos^{2} α$

$Find the Minimum Value$

$Conclusion$

The Sigma Insight: Properties of Inverse Trigonometric Functions

Analyzing the Setup

The Algebraic Dance

The Radical Trap

The Elegant Collapse

Similar Questions

Using the principal values of the inverse trigonometric functions, the sum of the maximum and the minimum values of $16 ((sec^{- 1} x)^{2} + (csc^{- 1} x)^{2})$ is:

Considering only the principal values of the inverse trigonometric functions, the value of $\frac{3}{2} cos^{- 1} \frac{2}{2 + π ^{2}} + \frac{1}{4} sin^{- 1} \frac{2 2 π}{2 + π ^{2}} + tan^{- 1} \frac{2}{π}$ is ______.

If $cos^{- 1} x - cos^{- 1} \frac{y}{2} = α$ , where $- 1 \leq x \leq 1, - 2 \leq y \leq 2, x \leq \frac{y}{2}$ , then for all $x, y, 4 x^{2} - 4 x y cos α + y^{2}$ is equal to

If $cos^{- 1} x - cos^{- 1} \frac{y}{2} = α$ , then $4 x^{2} - 4 x y cos α + y^{2}$ is equal to

If $sin^{- 1} (x - \frac{x ^{2}}{2} + \frac{x ^{3}}{4} - \dots) + cos^{- 1} (x^{2} - \frac{x ^{4}}{2} + \frac{x ^{6}}{4} - \dots) = \frac{π}{2}$ for $0 < ∣ x ∣ < 2$ , then $x$ equals

The sum of the absolute maximum and absolute minimum values of the function $f (x) = tan^{- 1} (sin x - cos x)$ in the interval $[0, π]$ is

Considering the principal values of the inverse trigonometric functions, $sin^{- 1} (\frac{3}{2} x + \frac{1}{2} 1 - x^{2}), - \frac{1}{2} < x < \frac{1}{2}$ , is equal to

Let $tan^{- 1} y = tan^{- 1} x + tan^{- 1} (\frac{2 x}{1 - x ^{2}})$ , where $∣ x ∣ < \frac{1}{3}$ . Then a value of $y$ is :

If $\frac{s i n ^{- 1} x}{a} = \frac{c o s ^{- 1} x}{b} = \frac{t a n ^{- 1} y}{c}$ ; $0 < x < 1$ , then the value of $cos (\frac{π c}{a + b})$ is:

Let the maximum value of $(sin^{- 1} x)^{2} + (cos^{- 1} x)^{2}$ for $x \in [- \frac{3}{2}, \frac{1}{2}]$ be $\frac{m}{n} π^{2}$ , where $gcd (m, n) = 1$ . Then $m + n$ is equal to ......... .

Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Analyze the Given Equation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Convert sin−1y to cos−1y

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Rearrange the Equation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Apply Addition Formula

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Take Cosine on Both Sides

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Isolate the Radical Term

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Square Both Sides

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Expand and Simplify

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Isolate the Target Expression

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Form: cos2α

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Find the Minimum Value

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Conclusion