( ⁻¹(3)/(5) + ⁻¹(5)/(13) + ⁻¹(33)/(65))$\cos \left(\sin^{-1}\frac{3}{5} +\sin^{-1}\frac{5}{13} +\sin^{-1}\frac{33}{65}\right)$ is equal to:

Related Formula

Apply the fundamental trigonometric identity properties directly linking reciprocal functions:

Core Logic

Simplify the given trigonometric equation using the identity formulas:

Substitute these simplified expressions back into the target relation equation:

Step 1: Set up a Quadratic Equation for the roots

We are given the linear \sum α + β = 8$\alpha + \beta = 8$. Use the algebraic identity for squares to find the product:

Since we know both the \sum (8) and product (15), α$\alpha$ and \β$\beta$ are the roots of the quadratic equation:

Step 2: Assign Variables and Compute the Target Value

Using the given constraint condition α ≤ β$\alpha \leq \beta$, we assign the values as:

Now substitute these values into the evaluation expression:

Pattern Recognition

Recognizing standard algebraic forms for sums and products like α+β$\alpha+\beta$ and αβ$\alpha\beta$ helps identify the system's values without needing to use full square root quadratic formulas.

Chapter Mix

Class 12 Mathematics: Inverse Trigonometric Functions

Related Formula

Core Logic

Rearrange using the principal trigonometric identity property ⁻¹ x = (π)/(2) - ⁻¹ x$\sin^{-1} x = \frac{\pi}{2} - \cos^{-1} x$:

Step 1: Isolate angles using variable substitution

Let ⁻¹ x = α$\cos^{-1} x = \alpha$ and ⁻¹(2x + 1) = β$\sin^{-1}(2x + 1) = \beta$.

Take cosine on both sides:

Step 2: Convert to algebraic identity form

Using double \angle formulas:

Since α = x$\cos\alpha = x$ and β = 2x + 1$\sin\beta = 2x + 1$, substitute directly:

Solving the quadratic equation gives:

Checking domain validation constraints for inverse functions dictates that only x = 1 - √(5)2$x = \frac{1 - \sqrt{5}}{2}$ is valid (the positive root exceeds principal bounds).

Step 3: Evaluate target question calculation

From the valid root, we track:

Squaring both sides yields:

Pattern Recognition

When inverse sums equate to values like (3π)/(2)$\frac{3\pi}{2}$, look for extreme boundary values or domain constraints to eliminate invalid algebraic roots.

Chapter Mix

Class 12 Mathematics: Inverse Trigonometric Functions Class 11 Mathematics: Trigonometric Equations

Related Formula

(2θ) = 1 - 2 ²θ$\cos(2\theta) = 1 - 2\sin^2\theta$

Core Logic

Let ⁻¹x = θ θ = x$\sin^{-1}x = \theta \implies \sin\theta = x$. The equation matches (2θ) = (1)/(9)$\cos(2\theta) = \frac{1}{9}$:

Since m$m$ and n$n$ are natural numbers, we pick x = (2)/(3) = (m)/(n)$x = \frac{2}{3} = \frac{m}{n}$. Because 2$2$ and 3$3$ are co-prime, we choose m = 2$m = 2$ and n = 3$n = 3$.

Step 1: Formulating Quadratic Equations

Substituting values into mx² - nx - m + n = 0$mx^2 - nx - m + n = 0$:

Factoring the equations:

Given α > β$\alpha > \beta$, we have α = 1$\alpha = 1$ and β = (1)/(2)$\beta = \frac{1}{2}$.

Step 2: Checking Options

Let us check the coordinates (1, (1)/(2))$\left(1, \frac{1}{2}\right)$ against option line configurations:

This exactly matches option (4).

Pattern Recognition

Co-prime conditions uniquely lock fractional values down to absolute integers. This bridges variables directly into standard algebraic calculations.

Chapter Mix

Class 12 Mathematics: Inverse Trigonometric Functions Class 10 Mathematics: Quadratic Equations

Related Formula

Core Logic

Evaluate a = ⁻¹( 5)$a = \sin^{-1}(\sin 5)$: The principal branch of ⁻¹ x$\sin^{-1} x$ is [-π/2, π/2]$[-\pi/2, \pi/2]$. 5$5$ radians is approximately 5 × 57.3^° ≈ 286.5^°$5 \times 57.3^\circ \approx 286.5^\circ$ (in 4th quadrant). The equivalent angle in the principal domain is 5 - 2π$5 - 2\pi$. Thus, a = 5 - 2π$a = 5 - 2\pi$.

Evaluate b = ⁻¹( 5)$b = \cos^{-1}(\cos 5)$: The principal branch of ⁻¹ x$\cos^{-1} x$ is [0, π]$[0, \pi]$. 5$5$ radians is in [π, 2π]$[\pi, 2\pi]$. The equivalent angle is 2π - 5$2\pi - 5$. Thus, b = 2π - 5$b = 2\pi - 5$.

Calculate a² + b²$a^2 + b^2$:

= 2(5 - 2π)²$= 2(5 - 2\pi)^2$

Chapter Mix

Class 12 Maths: Inverse Trigonometric Functions

Core Logic

Let ⁻¹α = A, ⁻¹β = B, ⁻¹γ = C$\sin^{-1}\alpha = A, \sin^{-1}\beta = B, \sin^{-1}\gamma = C$. Given A + B + C = π$A + B + C = \pi$. Since A = α, B = β, C = γ$\sin A = \alpha, \sin B = \beta, \sin C = \gamma$, α, β, γ$\alpha, \beta, \gamma$ act like the side lengths of a triangle divided by 2R$2R$ by Sine rule. However, directly dealing with the relation:

Step 1: Simplify Algebraic Relation

Step 2: Triangle Identification

Divide by 2αβ$2\alpha\beta$:

By Cosine Rule, C = (1)/(2)$\cos C = \frac{1}{2}$. Since C = ⁻¹γ$C = \sin^{-1}\gamma$, we know C = γ$\sin C = \gamma$. C = √(1 - γ²) = (1)/(2)$\cos C = \sqrt{1 - \gamma^2} = \frac{1}{2}$.

Step 3: Final Solution

Since C$C$ is an angle of a triangle (or sum equals π$\pi$ and elements are positive limits), γ = C > 0$\gamma = \sin C > 0$.

Pattern Recognition

The expression (α + β + γ)(α + β - γ) = 3αβ$(\alpha + \beta + \gamma)(\alpha + \beta - \gamma) = 3\alpha\beta$ perfectly mirrors the Cosine Rule standard form giving C = 1/2$\cos C = 1/2$.

Chapter Mix

Class 12 Maths: Inverse Trigonometric Functions Class 11 Maths: Trigonometric Functions