If cos⁻¹x - cos⁻¹(y/2) = α, where -1 ≤ x ≤ 1, -2 ≤ y ≤ 2, x ≤ y/2, then for all x, y, 4x² - 4xy cos

Q: Wait, why do we reject the negative root here?

In this problem, we are working with inverse cosine functions. The range of $\cos^{-1}x$ is $[0, \pi]$. When we take the cosine of both sides, we are essentially mapping the difference of two angles to a value. The squaring process is a standard algebraic technique to remove radicals, but it can introduce extraneous solutions. However, in this specific problem, the constraints $-1 \le x \le 1$ and $-2 \le y \le 2$ ensure that the values remain within the valid domain, and the algebraic manipulation leads us to a unique, consistent result.

Q: Why can't we just use energy conservation?

That is a great question! Energy conservation is a powerful tool in physics, but this is a pure mathematics problem involving inverse trigonometry. There is no physical system here to conserve energy for. We must rely on trigonometric identities and algebraic manipulation to solve it.

Q: Is there a domain restriction I should worry about?

Yes, absolutely! The problem explicitly states $-1 \le x \le 1$ and $-2 \le y \le 2$. These constraints are crucial because they ensure that the arguments of the inverse cosine functions are within their valid domains. Without these, the expressions $\cos^{-1}x$ and $\cos^{-1}\frac{y}{2}$ would be undefined. Always keep an eye on these constraints as they define the 'playground' where our math happens.

Question

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

Options:

(A) 4 \sin^2 \alpha - 2x^2y^2

(B) 4 \cos^2 \alpha + 2x^2y^2

(C) 4 \sin^2 \alpha

(D) 2 \sin^2 \alpha

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on the edge of a vast mathematical landscape. You have been given a problem that looks like a tangled mess of inverse trigonometric functions: $\cos^{-1}x - \cos^{-1}\frac{y}{2} = \alpha$.

Inverse trigonometry is simply a way of talking about angles. When you see $\cos^{-1}x$, view it as an angle $A$ whose cosine is $x$. Similarly, $\cos^{-1}\frac{y}{2}$ is an angle $B$ whose cosine is $\frac{y}{2}$.

The problem asks us to find the value of $4x^2 - 4xy \cos \alpha + y^2$. This expression is a quadratic form, completely devoid of inverse trigonometric functions. This is our target.

### The Cosine Bridge

To bridge the gap between inverse trigonometry and algebra, we take the cosine of both sides of the given equation:

$$\cos\left(\cos^{-1}x - \cos^{-1}\frac{y}{2}ight) = \cos \alpha$$

We now apply the compound angle identity $\cos(A-B) = \cos A \cos B + \sin A \sin B$. Substituting $A = \cos^{-1}x$ and $B = \cos^{-1}\frac{y}{2}$, we note that $\cos A = x$ and $\cos B = \frac{y}{2}$.

Using the Pythagorean identity $\sin^2 	heta + \cos^2 	heta = 1$, we determine the sine terms: $\sin A = \sqrt{1-x^2}$ and $\sin B = \sqrt{1-\frac{y^2}{4}}$. Substituting these into our identity yields:

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

### The Algebraic Dance

To eliminate the square roots, we isolate the radical term and square both sides:

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

Squaring both sides results in:

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

Expanding both sides, we obtain:

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

The term $\frac{x^2y^2}{4}$ appears on both sides and cancels out perfectly, leaving us with:

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

### The Final Flourish

To reach our target expression, we multiply the entire equation by $4$:

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

Rearranging the terms to group the variables on one side gives:

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

Since $4(1 - \cos^2 \alpha) = 4 \sin^2 \alpha$, we conclude that the value of the expression is:

**$4 \sin^2 \alpha$**

Animated Solution for Mathematics - Inverse Trigonometric Functions: 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

Select Answer:

Visualized Solution

The Given Equation

Applying the Cosine Function

The Substitution Step

Isolating the Radical

Squaring Both Sides

Expanding the Terms

Canceling and Simplifying

Multiplying by Four

Final Rearrangement

The Final Result

The Sigma Insight: Properties of Inverse Trigonometric Functions

Analyzing the Setup

The Cosine Bridge

The Algebraic Dance

The Final Flourish

Similar Questions

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

If $(sin^{- 1} x)^{2} - (cos^{- 1} x)^{2} = a; 0 < x < 1, a \neq = 0$ , then the value of $2 x^{2} - 1$ is :

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

Prove that $cos tan^{- 1} sin cot^{- 1} x = \frac{x ^{2} + 1}{x ^{2} + 2}$ .

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

If $\frac{π}{2} \leq x \leq \frac{3 π}{4}$ then $cos^{- 1} (\frac{12}{13} cos x + \frac{5}{13} sin x)$ is equal to

If the sum of all the solutions of $tan^{- 1} (\frac{2 x}{1 - x ^{2}}) + cot^{- 1} (\frac{1 - x ^{2}}{2 x}) = \frac{π}{3}$ , $- 1 < x < 1, x \neq = 0$ , is $α - \frac{4}{3}$ , then $α$ is equal to

$1 + x^{2} [{x cos (cot^{- 1} x) + sin (cot^{- 1} x)}^{2} - 1]^{1/2} =$

If for some $α, β \in R$ such that $α \leq β$ , $α + β = 8$ and $sec^{2} (tan^{- 1} α) + csc^{2} (cot^{- 1} β) = 36,$ then $α^{2} + β$ is equal to