If cos⁻¹ x - cos⁻¹ (y/2) = α, then 4x² - 4xy cos α + y² is equal to (A) 2 sin 2α (B) 4 (C) 4 sin² α

Q: Wait, why can't I just take the cosine of both sides immediately and write $\cos(\cos^{-1} x) - \cos(\cos^{-1} \frac{y}{2}) = \cos \alpha$?

This is a classic trap! Remember, the cosine function is not linear. That is, $\cos(A - B)$ is NOT equal to $\cos A - \cos B$. You cannot distribute the cosine function across a subtraction. You must use the compound angle formula $\cos(A - B) = \cos A \cos B + \sin A \sin B$. By using the inverse identity, we are essentially applying this compound angle formula in reverse, which is the only mathematically sound way to proceed.

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

You *could* square immediately, but it would be a tactical error. If you square the expression $\frac{xy}{2} + \sqrt{1-x^2}\sqrt{1-\frac{y^2}{4}}$ directly, you will get a cross-product term that still contains the square root, meaning you haven't actually eliminated the radical. By isolating the radical term on one side, you ensure that when you square, the radical disappears completely, leaving you with a clean polynomial equation. It is all about efficiency and avoiding unnecessary complexity.

Q: Is there a domain restriction I should worry about for $x$ and $y$?

In the context of JEE problems like this, we generally assume the variables lie within the principal domain of the inverse cosine function, meaning $x \in [-1, 1]$ and $\frac{y}{2} \in [-1, 1]$. While it is good practice to keep this in mind, the algebraic manipulation we performed is valid regardless of the specific values, provided the expression is defined. Focus on the algebraic beauty of the identity first!

Question

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

Options:

(A) 2 \sin 2\alpha

(B) 4

(C) 4 \sin^2 \alpha

(D) - 4 \sin^2 \alpha

Sigma Percentile · Accepted Answer

### Analyzing the Setup

My dear student, welcome to the arena. Today, we are going to dismantle a problem that often intimidates students because it blends two worlds: the abstract, often confusing world of inverse trigonometry and the structured, predictable world of algebra.

When you see an equation like $\cos^{-1} x - \cos^{-1} \frac{y}{2} = \alpha$, your instinct might be to panic. But I want you to take a deep breath. In the JEE Advanced, these problems are not designed to break you; they are designed to test your ability to see the underlying structure.

Let us peel back the layers together.

### The Identity as a Bridge

We are given $\cos^{-1} x - \cos^{-1} \frac{y}{2} = \alpha$. Our target is to find the value of $4x^2 - 4xy \cos \alpha + y^2$.

Notice the disconnect? The given is in 'inverse-trig land,' and the target is in 'algebra land.' We need a bridge. That bridge is the identity for the difference of two inverse cosines:

$$\cos^{-1} A - \cos^{-1} B = \cos^{-1} (AB + \sqrt{1-A^2}\sqrt{1-B^2})$$

Why does this work? Think of it as the cosine subtraction formula, $\cos(A-B) = \cos A \cos B + \sin A \sin B$, wearing a disguise. When we set $A = x$ and $B = \frac{y}{2}$, we are essentially converting the angles back into their trigonometric ratios.

By substituting these into our identity, we transform our equation into:

$$\cos^{-1} \left( x \cdot \frac{y}{2} + \sqrt{1-x^2}\sqrt{1-\frac{y^2}{4}} \right) = \alpha$$

Now, the path forward becomes clear. We want to strip away the $\cos^{-1}$ function by taking the cosine of both sides. The inverse function vanishes, and we are left with the raw algebraic core:

$$\frac{xy}{2} + \sqrt{1-x^2}\sqrt{1-\frac{y^2}{4}} = \cos \alpha$$

### The Art of Algebraic Surgery

Now, we face the radicals. Many students rush here, squaring both sides immediately. Do not do that!

If you square the left side as it stands, you will create a cross-product term that keeps the square root alive, and you will be back to square one. We must perform 'algebraic surgery' by isolating the radical term:

$$\sqrt{1-x^2}\sqrt{1-\frac{y^2}{4}} = \cos \alpha - \frac{xy}{2}$$

Now, and only now, do we square both sides. This is the decisive move. On the left, the radicals vanish, leaving us with a simple product.

On the right, we expand the binomial $(\cos \alpha - \frac{xy}{2})^2$. Let us watch the magic unfold:

$$(1-x^2)\left(1-\frac{y^2}{4}\right) = \cos^2 \alpha - 2(\cos \alpha)\left(\frac{xy}{2}\right) + \frac{x^2y^2}{4}$$

### The Elegant Cancellation

Expand the left side carefully. We get $1 - \frac{y^2}{4} - x^2 + \frac{x^2y^2}{4}$.

Now, look at the right side. The term $-2(\cos \alpha)(\frac{xy}{2})$ simplifies beautifully to $-xy \cos \alpha$. So our equation is:

$$1 - \frac{y^2}{4} - x^2 + \frac{x^2y^2}{4} = \cos^2 \alpha - xy \cos \alpha + \frac{x^2y^2}{4}$$

Do you see it? The term $\frac{x^2y^2}{4}$ appears on both sides. It cancels out perfectly! This is the moment where the problem stops being a chore and starts being a symphony.

We are left with:

$$1 - \frac{y^2}{4} - x^2 = \cos^2 \alpha - xy \cos \alpha$$

### Final Calculation

We are almost home. We need to match our target expression: $4x^2 - 4xy \cos \alpha + y^2$.

Currently, we have fractions and negative signs. Let us multiply the entire equation by $4$ to clear the denominators:

$$4 - y^2 - 4x^2 = 4 \cos^2 \alpha - 4xy \cos \alpha$$

Now, rearrange the terms to isolate the target expression on one side. Move $-4x^2$ and $-y^2$ to the right, and move $4 \cos^2 \alpha$ to the left:

$$4 - 4 \cos^2 \alpha = 4x^2 - 4xy \cos \alpha + y^2$$

Factor out the $4$ on the left side, and we arrive at the final, beautiful identity:

$$4(1 - \cos^2 \alpha) = 4x^2 - 4xy \cos \alpha + y^2$$

Since $1 - \cos^2 \alpha = \sin^2 \alpha$, our final answer is simply **$4 \sin^2 \alpha$**.

Animated Solution for Mathematics - Inverse Trigonometric Functions: If $cos^{- 1} x - cos^{- 1} \frac{y}{2} = α$ , then $4 x^{2} - 4 x y cos α + y^{2}$ is equal to

Select Answer:

Visualized Solution

The Given Equation

The Inverse Cosine Formula

Applying the Formula

Eliminating Inverse Cosine

Isolating the Square Roots

Squaring Both Sides

Expanding the Equation

Simplifying the Equation

Clearing the Denominators

Rearranging to the Target Expression

The Final Answer

The Sigma Insight: Properties of Inverse Trigonometric Functions

Analyzing the Setup

The Identity as a Bridge

The Art of Algebraic Surgery

The Elegant Cancellation

Final Calculation

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