Considering the principal values of the inverse trigonometric functions, the sum of all the solution

Q: Wait, why do we need to check the domain? Can't we just solve the algebra?

That is a great question! In inverse trigonometry, algebraic manipulation often introduces 'extraneous roots'—solutions that satisfy the final polynomial but violate the original domain constraints. Because $\cos^{-1}(2x)$ is only defined for $x \in [-\frac{1}{2}, \frac{1}{2}]$, any root outside this interval is mathematically invalid. Always check your domain first!

Q: Why does $\cos(3\theta - \pi)$ become $-\cos(3\theta)$?

This comes from the property of allied angles. We know that $\cos(\theta - \pi) = \cos(-(\pi - \theta))$. Since the cosine function is even, $\cos(-x) = \cos(x)$, so this is equal to $\cos(\pi - \theta)$. Since $\cos(\pi - \theta)$ is in the second quadrant, it simplifies to $-\cos(\theta)$. It is a classic step, so keep it in your toolkit!

Q: Do I really need to verify every root? It takes so much time.

I understand the frustration, but in JEE Advanced, verification is not just a safety net—it is part of the problem's design. Especially with inverse trigonometric equations, the principal value branches can be tricky. Verifying the roots ensures you haven't accidentally included a value that falls outside the principal range. It is the final, disciplined act of a true mathematician.

Question

Considering the principal values of the inverse trigonometric functions, the sum of all the solutions of the equation $\cos^{-1}(x) - 2\sin^{-1}(x) = \cos^{-1}(2x)$ is equal to :

Options:

(A) 0

(B) 1

(C) $\frac{1}{2}$

(D) $-\frac{1}{2}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

The equation is given by $\cos^{-1}(x) - 2\sin^{-1}(x) = \cos^{-1}(2x)$. Before proceeding with algebraic manipulation, we must define the domain of validity for the variables.

The functions $\cos^{-1}(x)$ and $\sin^{-1}(x)$ are defined for $x \in [-1, 1]$. However, the term $\cos^{-1}(2x)$ imposes a stricter constraint:
$$ -1 \le 2x \le 1 \implies x \in \left[ -\frac{1}{2}, \frac{1}{2} ight] $$

This interval represents the "soil" of our problem, the only region where potential solutions can exist.

### The Unification

To solve the equation, we bridge the gap between $\cos^{-1}$ and $\sin^{-1}$ using the identity:
$$ \sin^{-1}(x) + \cos^{-1}(x) = \frac{\pi}{2} $$

By substituting $\sin^{-1}(x) = \frac{\pi}{2} - \cos^{-1}(x)$ into the original equation, we transform it into a single language:
$$ \cos^{-1}(x) - 2\left( \frac{\pi}{2} - \cos^{-1}(x) ight) = \cos^{-1}(2x) $$

Expanding this expression yields:
$$ \cos^{-1}(x) - \pi + 2\cos^{-1}(x) = \cos^{-1}(2x) $$
$$ 3\cos^{-1}(x) - \pi = \cos^{-1}(2x) $$

### The Algebraic Leap

To isolate $x$, we take the cosine of both sides:
$$ \cos(3\cos^{-1}(x) - \pi) = \cos(\cos^{-1}(2x)) $$

The right side simplifies directly to $2x$. For the left side, we use the allied angle property $\cos(	heta - \pi) = -\cos(	heta)$, where $	heta = 3\cos^{-1}(x)$:
$$ -\cos(3\cos^{-1}(x)) = 2x $$

Applying the triple angle formula $\cos(3\alpha) = 4\cos^3(\alpha) - 3\cos(\alpha)$ with $\alpha = \cos^{-1}(x)$, we substitute $\cos(\alpha) = x$:
$$ -(4x^3 - 3x) = 2x $$
$$ 4x^3 - x = 0 $$

### The Final Gatekeeper

Factoring the polynomial $x(4x^2 - 1) = 0$ provides three potential candidates:
$$ x = 0, \quad x = \frac{1}{2}, \quad x = -\frac{1}{2} $$

We verify these against our domain $x \in [-\frac{1}{2}, \frac{1}{2}]$:

1. For $x = 0$: $\cos^{-1}(0) - 2\sin^{-1}(0) = \frac{\pi}{2} - 0 = \frac{\pi}{2}$, and $\cos^{-1}(0) = \frac{\pi}{2}$. (Valid)
2. For $x = \frac{1}{2}$: $\cos^{-1}(\frac{1}{2}) - 2\sin^{-1}(\frac{1}{2}) = \frac{\pi}{3} - 2(\frac{\pi}{6}) = 0$, and $\cos^{-1}(1) = 0$. (Valid)
3. For $x = -\frac{1}{2}$: $\cos^{-1}(-\frac{1}{2}) - 2\sin^{-1}(-\frac{1}{2}) = \frac{2\pi}{3} - 2(-\frac{\pi}{6}) = \pi$, and $\cos^{-1}(-1) = \pi$. (Valid)

All three values are valid solutions. The sum of these solutions is:
$$ 0 + \frac{1}{2} + \left( -\frac{1}{2} ight) = \mathbf{0} $$

Animated Solution for Mathematics - Inverse Trigonometric Functions: Considering the principal values of the inverse trigonometric functions, the sum of all the solutions of the equation $cos^{- 1} (x) - 2 sin^{- 1} (x) = cos^{- 1} (2 x)$ is equal to :

Select Answer:

Visualized Solution

Domain Constraints

Inverse Trigonometric Identity

Substitution

Simplification

Taking Cosine

Cosine Transformation

Triple Angle Formula

Algebraic Equation

Factorization

Verification: $x = 0$

Verification: $x = \frac{1}{2}$

Verification: $x = - \frac{1}{2}$

Sum of Solutions

The Sigma Insight: Solving Inverse Trigonometric Equations

Analyzing the Setup

The Unification

The Algebraic Leap

The Final Gatekeeper

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