Let the maximum value of (sin⁻¹ x)² + (cos⁻¹ x)² for x ∈ [-√3/2, 1/√2] be (m/n)π², where gcd(m, n) =

Q: Wait, why can't we just find the maximum of $(\sin^{-1} x)^2$ and $(\cos^{-1} x)^2$ separately and add them?

That is a classic trap! The variables $\sin^{-1} x$ and $\cos^{-1} x$ are not independent; they are linked by the identity $\sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}$. Because they are constrained by the same $x$, you must treat the expression as a single function of $x$ (or $t$) rather than two separate components.

Q: Why is the maximum at the endpoint and not the vertex of the parabola?

Great observation! The quadratic $f(t) = 2t^2 - \pi t + \frac{\pi^2}{4}$ is an upward-opening parabola (since the coefficient of $t^2$ is positive). For an upward-opening parabola, the vertex is the *minimum* point. The maximum value on a closed interval will always occur at the endpoint that is furthest from the vertex.

Q: How do I determine the range of $t = \sin^{-1} x$?

Since $t = \sin^{-1} x$, the range of $t$ is simply the interval $[\sin^{-1}(x_{min}), \sin^{-1}(x_{max})]$. Given $x \in [-\frac{\sqrt{3}}{2}, \frac{1}{\sqrt{2}}]$, we calculate $\sin^{-1}(-\frac{\sqrt{3}}{2}) = -\frac{\pi}{3}$ and $\sin^{-1}(\frac{1}{\sqrt{2}}) = \frac{\pi}{4}$. Thus, $t \in [-\frac{\pi}{3}, \frac{\pi}{4}]$.

Question

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

Sigma Percentile · Accepted Answer

### Analyzing the Setup

The expression given is $f(x) = (\sin^{-1} x)^2 + (\cos^{-1} x)^2$ within the interval $x \in [-\frac{\sqrt{3}}{2}, \frac{1}{\sqrt{2}}]$.

To simplify this, we utilize the fundamental identity:
$$\sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}$$

Let $t = \sin^{-1} x$. Consequently, $\cos^{-1} x = \frac{\pi}{2} - t$. The expression transforms into a function of $t$:
$$f(t) = t^2 + \left( \frac{\pi}{2} - t ight)^2$$

Expanding this expression yields:
$$f(t) = t^2 + \frac{\pi^2}{4} - \pi t + t^2 = 2t^2 - \pi t + \frac{\pi^2}{4}$$

### The Geometry of the Quadratic

We have arrived at a quadratic function $f(t) = 2t^2 - \pi t + \frac{\pi^2}{4}$. Since the coefficient of $t^2$ is positive, this represents an upward-opening parabola.

To identify the vertex, we complete the square:
$$f(t) = 2\left( t^2 - \frac{\pi}{2}t ight) + \frac{\pi^2}{4}$$
$$f(t) = 2\left( t - \frac{\pi}{4} ight)^2 + \frac{\pi^2}{8}$$

The vertex of this parabola is at $t = \frac{\pi}{4}$, where the function attains its absolute minimum value of $\frac{\pi^2}{8}$.

### The Constraint of the Domain

We must respect the restricted domain $x \in [-\frac{\sqrt{3}}{2}, \frac{1}{\sqrt{2}}]$. Mapping these values through $t = \sin^{-1} x$, we find the interval for $t$:
$$t \in \left[ \sin^{-1}\left( -\frac{\sqrt{3}}{2} ight), \sin^{-1}\left( \frac{1}{\sqrt{2}} ight) ight] = \left[ -\frac{\pi}{3}, \frac{\pi}{4} ight]$$

The vertex $t = \frac{\pi}{4}$ lies at the right boundary of our interval. Because the parabola opens upward, the maximum value must occur at the point furthest from the vertex.

Comparing the distances, the left endpoint $t = -\frac{\pi}{3}$ is significantly further from the vertex than the right endpoint. Therefore, the maximum occurs at $t = -\frac{\pi}{3}$.

### The Final Calculation

Substituting $t = -\frac{\pi}{3}$ into our quadratic expression:
$$f\left( -\frac{\pi}{3} ight) = 2\left( -\frac{\pi}{3} ight)^2 - \pi\left( -\frac{\pi}{3} ight) + \frac{\pi^2}{4}$$

$$f\left( -\frac{\pi}{3} ight) = \frac{2\pi^2}{9} + \frac{\pi^2}{3} + \frac{\pi^2}{4}$$

Finding a common denominator of $36$:
$$f\left( -\frac{\pi}{3} ight) = \frac{8\pi^2}{36} + \frac{12\pi^2}{36} + \frac{9\pi^2}{36} = \frac{29\pi^2}{36}$$

Given the form $\frac{m}{n}\pi^2$, we identify $m = 29$ and $n = 36$. Since $	ext{gcd}(29, 36) = 1$, the final result is:
**$m + n = 29 + 36 = 65$**

Animated Solution for Mathematics - Inverse Trigonometric Functions: 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 ......... .

Enter Numerical Value:

Visualized Solution

Introduction to $f (x)$

The Identity $sin^{- 1} x + cos^{- 1} x = \frac{π}{2}$

Substituting and Expanding

Forming the Quadratic $f (t)$

Completing the Square

Identifying the Vertex

Determining the Domain of $t$

Visualizing the Interval

Finding the Maximum Point

Calculating $f (- \frac{π}{3})$

Summing the Fractions

Identifying $m$ and $n$

Final Result: $m + n$

The Sigma Insight: Properties of Inverse Trigonometric Functions

Analyzing the Setup

The Geometry of the Quadratic

The Constraint of the Domain

The Final Calculation

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Enter Numerical Value:

Visualized Solution

Introduction to .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x)

The Identity .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } sin−1x+cos−1x=2π​

Substituting and Expanding

Forming the Quadratic .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(t)

Completing the Square

Identifying the Vertex

Determining the Domain of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } t

Visualizing the Interval

Finding the Maximum Point

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(−3π​)

Summing the Fractions

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

Final Result: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } m+n