If S = {x ∈ R : sin⁻¹((x+1)/√(x²+2x+2)) - sin⁻¹(x/√(x²+1)) = π/4} then Σ_{x∈S} (sin((x² + x + 5)π/2)

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

Actually, we don't! Both $x=0$ and $x=-1$ are valid solutions. Always check the domain of the inverse sine function, but here, the arguments are always within $[-1, 1]$, so both are perfectly valid. Don't let the negative sign scare you; it's just a number on the number line!

Q: Why can't we just use the identity $\sin^{-1} A - \sin^{-1} B = \sin^{-1}(A\sqrt{1-B^2} - B\sqrt{1-A^2})$?

You absolutely could, but that path is a minefield of algebraic complexity! You would end up with nested square roots and a much higher probability of making a calculation error. The geometric substitution method is the 'elegant' path—it transforms a complex trigonometric equation into a simple algebraic one. In JEE Advanced, choosing the elegant path is often the difference between finishing on time and getting stuck.

Q: How did you know $x^2+x+5$ would be the same for both roots?

That is the beauty of symmetry! The quadratic expression $x^2+x$ is symmetric about its vertex at $x = -0.5$. Since our roots $0$ and $-1$ are equidistant from $-0.5$ (both are $0.5$ units away), the value of the quadratic expression must be identical for both. Always look for these patterns before diving into heavy arithmetic!

Question

If $S = \{x \in \mathbb{R} : \sin^{-1}\left(\frac{x+1}{\sqrt{x^2+2x+2}}\right) - \sin^{-1}\left(\frac{x}{\sqrt{x^2+1}}\right) = \frac{\pi}{4}\}$ then $\sum_{x \in S} \left(\sin\left((x^2+x+5)\frac{\pi}{2}\right) - \cos((x^2+x+5)\pi)\right)$ is equal to _________.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

The given equation is:

$$\sin^{-1}\left(\frac{x+1}{\sqrt{x^2+2x+2}}ight) - \sin^{-1}\left(\frac{x}{\sqrt{x^2+1}}ight) = \frac{\pi}{4}$$

At first glance, this appears complex. However, observe the denominators. The first denominator is $\sqrt{x^2+2x+2}$, which simplifies to $\sqrt{(x+1)^2 + 1}$ by completing the square. The second denominator is $\sqrt{x^2+1}$.

### The Geometric Lens

Recall that in a right-angled triangle with perpendicular $u$ and base $1$, the hypotenuse is $\sqrt{u^2+1}$. Consequently, $\sin 	heta = \frac{u}{\sqrt{u^2+1}}$.

The terms inside our inverse sine functions follow this exact structure. Let $	an \alpha = x+1$ and $	an \beta = x$. Then:

$$\sin \alpha = \frac{x+1}{\sqrt{(x+1)^2+1}} \quad 	ext{and} \quad \sin \beta = \frac{x}{\sqrt{x^2+1}}$$

This transforms our equation into a simple relationship between angles: $\alpha - \beta = \frac{\pi}{4}$.

### The Algebraic Collapse

To solve for $x$, we apply the tangent function to both sides of the equation $\alpha - \beta = \frac{\pi}{4}$:

$$	an(\alpha - \beta) = 	an\left(\frac{\pi}{4}ight)$$

Using the identity $	an(\alpha - \beta) = \frac{	an \alpha - 	an \beta}{1 + 	an \alpha 	an \beta}$ and knowing $	an(\frac{\pi}{4}) = 1$, we substitute our expressions for $	an \alpha$ and $	an \beta$:

$$\frac{(x+1) - x}{1 + (x+1)x} = 1$$

The numerator simplifies to $1$, yielding the equation $\frac{1}{1 + x^2 + x} = 1$. Cross-multiplying gives $1 + x^2 + x = 1$, which simplifies to $x^2 + x = 0$. Factoring this, we find the roots: **$x = 0$** and **$x = -1$**.

### Final Calculation

We must evaluate the expression $\sin((x^2+x+5)\frac{\pi}{2}) - \cos((x^2+x+5)\pi)$ for $x \in \{0, -1\}$. Note that for both roots, the core quadratic $x^2+x+5$ evaluates to $5$.

Substituting $5$ into the expression:

$$\sin\left(5 \cdot \frac{\pi}{2}ight) - \cos(5\pi)$$

We know that $\sin(\frac{5\pi}{2}) = \sin(2\pi + \frac{\pi}{2}) = 1$ and $\cos(5\pi) = \cos(4\pi + \pi) = -1$. Thus, the value for each root is:

$$1 - (-1) = 2$$

Since the expression evaluates to $2$ for both $x=0$ and $x=-1$, the total sum is $2 + 2 = 4$. The final answer is **$4$**.

Animated Solution for Mathematics - Inverse Trigonometric Functions: If $S = {x \in R : sin^{- 1} (\frac{x + 1}{x ^{2} + 2 x + 2}) - sin^{- 1} (\frac{x}{x ^{2} + 1}) = \frac{π}{4}}$ then $\sum_{x \in S} (sin ((x^{2} + x + 5) \frac{π}{2}) - cos ((x^{2} + x + 5) π))$ is equal to _________.

Enter Numerical Value:

Visualized Solution

Analyze the Equation Structure

Constructing Triangle $α$

Constructing Triangle $β$

Simplify the Equation

Apply Tangent on Both Sides

Substitute $x$ Back

Solve the Quadratic Equation

Analyze the Target Expression

Evaluate the Core Term

Calculate the Expression Value

Final Summation

The Sigma Insight: Solving Inverse Trigonometric Equations

Analyzing the Setup

The Geometric Lens

The Algebraic Collapse

Final Calculation

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