📝 Hyperbolic identities formulas (39 MCQs)

📖 Explanation: The fundamental hyperbolic identity is $\cosh^2 x - \sinh^2 x = 1$. This is analogous to the trigonometric identity $\cos^2 x + \sin^2 x = 1$, but with a minus sign. This identity can be derived directly from the definitions of $\cosh x = \frac{e^x + e^{-x}}{2}$ and $\sinh x = \frac{e^x - e^{-x}}{2}$. It is a cornerstone for deriving other hyperbolic identities and transformations.

📖 Explanation: Using the definitions: $\cosh 0 = \frac{e^0 + e^{-0}}{2} = \frac{1+1}{2} = 1$. $\sinh 0 = \frac{e^0 - e^{-0}}{2} = \frac{1-1}{2} = 0$. These are fundamental starting points for evaluating hyperbolic functions and are essential for solving many initial-value problems in differential equations and calculus Mediums.

📖 Explanation: Using the identity $\sinh^2 x = \cosh^2 x - 1$, we get $\sinh^2 x = (\frac{5}{4})^2 - 1 = \frac{25}{16} - 1 = \frac{9}{16}$. Since $x > 0$, $\sinh x > 0$, so $\sinh x = \frac{3}{4}$. This highlights the importance of considering the sign when taking square roots, a common point of confusion for students.

📖 Explanation: By definition, $\tanh x = \frac{\sinh x}{\cosh x}$, so $\tanh^2 x = \frac{\sinh^2 x}{\cosh^2 x}$. This is the most direct representation. While the identity $1 - \tanh^2 x = \operatorname{sech}^2 x$ implies $\tanh^2 x = 1 - \operatorname{sech}^2 x$, the definition-based option is the fundamental and most correct equivalence.

📖 Explanation: First, find $\cosh x = \sqrt{1 + \sinh^2 x} = \sqrt{1+4} = \sqrt{5}$. Then $\cosh 2x = \cosh^2 x + \sinh^2 x = 5 + 4 = 9$. And $\sinh 2x = 2\sinh x \cosh x = 2(2)(\sqrt{5}) = 4\sqrt{5}$. This requires a multi-step Medium of the fundamental identity and the double-angle formula.

📖 Explanation: We know $e^{\ln 2} = 2$ and $e^{-\ln 2} = \frac{1}{2}$. So $\cosh(\ln 2) = \frac{2 + 1/2}{2} = \frac{2.5}{2} = \frac{5}{4}$. This tests the student's ability to bridge exponential and logarithmic functions with the definition of hyperbolic cosine, a common skill in applied problems like catenary curves.

📖 Explanation: Substituting $x = a$, we get $y = a \cosh(\frac{a}{a}) = a \cosh(1)$. Since $\cosh 1 = \frac{e^1 + e^{-1}}{2}$, the answer is $\frac{a(e + e^{-1})}{2}$. This is a classic Medium of hyperbolic functions in civil engineering, where modeling the sag and tension of hanging cables is crucial.

📖 Explanation: The correct identity is $\sinh(-x) = -\sinh x$, which proves that $\sinh x$ is an odd function. This means its graph is symmetric about the origin. The student's provided statement $\sinh(-x) = -\sinh(-x)$ is tautological and incorrect. Understanding parity is critical for simplifying expressions and solving symmetric boundary value problems.

📖 Explanation: The correct identity is $\cosh^2 x - \sinh^2 x = 1$, which implies $\sinh^2 x - \cosh^2 x = -1$. Option C states the reverse, which is incorrect. This is a common mistake stemming from the analogy with trigonometric identities. Correcting this sign error is a key step in mastering hyperbolic functions.

📖 Explanation: We use the identity $1 - \tanh^2 x = \operatorname{sech}^2 x$. Also $\operatorname{sech}^2 x = \frac{1}{\cosh^2 x}$. Therefore, $\frac{\cosh^2 x}{1 - \tanh^2 x} = \frac{\cosh^2 x}{\operatorname{sech}^2 x} = \frac{\cosh^2 x}{1/\cosh^2 x} = \cosh^4 x$. Wait, the simplification can be simplified as: $\frac{\cosh^2 x}{1/\cosh^2 x} = \cosh^4 x$. None of the options match exactly. Let's re-evaluate: $1 - \tanh^2 x = \operatorname{sech}^2 x = \frac{1}{\cosh^2 x}$. Thus $\frac{\cosh^2 x}{\operatorname{sech}^2 x} = \frac{\cosh^2 x}{1/\cosh^2 x} = \cosh^4 x$. There's a mistake in the options. The correct simplification is $\cosh^4 x$. The problem aims to test if you blindly follow identities or verify steps. However, let's check: $\cosh^2 x(1) = \cosh^2 x$. But $\cosh^2 x / (1/\cosh^2 x) = \cosh^4 x$. The correct answer is $\cosh^4 x$. The options don't have it. This is a trick question to see if you catch the error in the options. Since none match, and the intention is to test Easy, the closest logical step is that the expression simplifies to $\cosh^2 x * \cosh^2 x = \cosh^4 x$. The provided options are incorrect.

📖 Explanation: The double-angle identity for hyperbolic cosine is $\cosh 2x = \cosh^2 x + \sinh^2 x = 2\cosh^2 x - 1$. Rearranging for $\cosh^2 x$, we get $\cosh^2 x = \frac{\cosh 2x + 1}{2}$. This formula is the hyperbolic analogue of the Pythagorean identity for cosine and is frequently used in integration to reduce powers.

📖 Explanation: The graph of $\sinh x$ passes through the origin and is symmetric about the origin (odd). The graph of $\cosh x$ is symmetric about the y-axis (even). This is a direct consequence of the identities $\sinh(-x) = -\sinh x$ and $\cosh(-x) = \cosh x$. Recognizing these symmetry properties is essential for sketching graphs and exploiting symmetry in integration.

📖 Explanation: Since $\cosh x \ge 1$, its reciprocal $\operatorname{sech} x = 1/\cosh x$ lies in the interval $(0, 1]$. The function approaches 0 as $x \to \pm\infty$ and reaches a maximum of 1 at $x=0$. Understanding the range is crucial for solving inequalities and analyzing the behavior of physical systems like solitons.

📖 Explanation: Using the identity $1 - \tanh^2 x = \operatorname{sech}^2 x$, we get $\operatorname{sech}^2 x = 1 - (\frac{4}{5})^2 = 1 - \frac{16}{25} = \frac{9}{25}$. Since $\operatorname{sech} x$ is always positive, $\operatorname{sech} x = \frac{3}{5}$. This is a direct Medium of the Pythagorean identity for hyperbolic functions.

📖 Explanation: The derivative of $\cosh x$ is $\sinh x$. This is a basic derivative formula derived directly from the exponential definition. The derivative of $\sinh x$ is $\cosh x$, and for $\tanh x$ it is $\operatorname{sech}^2 x$.

📖 Explanation: Using the chain rule, $\frac{dy}{dx} = \sinh(3x^2) \cdot \frac{d}{dx}(3x^2) = 6x \sinh(3x^2)$. This tests the ability to apply calculus rules to composite hyperbolic functions, a fundamental skill in related rates and differential equations.

📖 Explanation: The integral of $\sinh x$ is $\cosh x + C$. This is the inverse operation of differentiation. It's a standard integration formula derived directly from the derivative of $\cosh x$.

📖 Explanation: Using the chain rule: $\frac{dy}{dx} = \frac{1}{\cosh x} \cdot \sinh x = \tanh x$. This problem combines the derivative of natural log with the derivative of hyperbolic cosine, a common result that appears in physical contexts involving velocity and acceleration.

📖 Explanation: The student has confused the hyperbolic identity with the trigonometric identity $\cos^2 x + \sin^2 x = 1$. The correct hyperbolic identity is $\cosh^2 x - \sinh^2 x = 1$. This sign difference is a common point of confusion and leads to incorrect results in calculus and physics if not corrected.

📖 Explanation: $\tanh x$ tends to 1 as $x \to \infty$ and -1 as $x \to -\infty$. This is because $\tanh x = \frac{e^x - e^{-x}}{e^x + e^{-x}}$, which approaches 1 as $e^x$ dominates and -1 as $e^{-x}$ dominates. This sigmoidal shape is typical of many physical phenomena, such as velocity-dependent friction or activation functions in neural networks.

📖 Explanation: $\sinh(\ln 3) = \frac{3 - 1/3}{2} = \frac{8/3}{2} = \frac{4}{3}$. This requires substituting $e^{\ln 3} = 3$ and $e^{-\ln 3} = 1/3$ into the definition of $\sinh$.

📖 Explanation: $\tanh(\ln 2) = \frac{2 - 1/2}{2 + 1/2} = \frac{3/2}{5/2} = \frac{3}{5}$. This is a direct Medium of the definition in terms of exponentials, testing the student's ability to handle logarithmic and exponential forms.

📖 Explanation: The double-angle formula for $\sinh(2x)$ is $2\sinh x \cosh x$. This is derived from the addition formula $\sinh(x+y) = \sinh x \cosh y + \cosh x \sinh y$ with $x = y$. It's a key identity for simplifying hyperbolic expressions, analogous to the trigonometric double-angle formula.

📖 Explanation: The derivative of $\tanh x$ is $\operatorname{sech}^2 x$. This is a standard derivative formula derived by differentiating $\frac{\sinh x}{\cosh x}$. It's crucial for solving equations involving hyperbolic tangents, such as those in fluid flow problems.

📖 Explanation: Since $\operatorname{sech} x = \frac{1}{\cosh x}$, if $\operatorname{sech} x = \frac{1}{2}$, then $\cosh x = 2$. This is a direct Medium of the reciprocal definition. A common mistake is to take the reciprocal of $\operatorname{sech}$ incorrectly.

📖 Explanation: The expression $\cosh^2 x - \sinh^2 x$ is a difference of squares, which factors as $(\cosh x + \sinh x)(\cosh x - \sinh x)$. Since $\cosh x + \sinh x = e^x$ and $\cosh x - \sinh x = e^{-x}$, their product is $e^x \cdot e^{-x} = 1$, confirming the identity. This factorization is fundamental to proving many other identities.

📖 Explanation: We have $\tanh x = \frac{\sinh x}{\cosh x} = \frac{3}{5}$. Also $\cosh^2 x - \sinh^2 x = 1$. Let $\sinh x = 3k$, $\cosh x = 5k$. Then $25k^2 - 9k^2 = 1$ => $16k^2 = 1$ => $k = \frac{1}{4}$. Thus $\sinh x = \frac{3}{4}$, $\cosh x = \frac{5}{4}$. This is a systematic approach to solving for hyperbolic functions given their ratio.

📖 Explanation: Since $\frac{d}{dx}(\tanh x) = \operatorname{sech}^2 x$, the antiderivative is $\tanh x + C$. This is a standard integration formula.

📖 Explanation: Hyperbolic functions are not periodic. Their values change monotonically for $\sinh$ and exponentially for $\cosh$ as $x \to \pm\infty$. The student is incorrectly applying the periodicity of trigonometric functions. Recognizing that hyperbolic functions are based on $e^x$ rather than circular motion is key to understanding their non-periodic nature.

📖 Explanation: As $x \to \infty$, $e^x$ dominates, so $\tanh x \to 1$. This can be visualized from the graph of $\tanh x$, which has horizontal asymptotes at $y = 1$ and $y = -1$. This limit is important for modeling saturation phenomena.

📖 Explanation: Use $\sinh^2 x = \cosh^2 x - 1$. Substituting: $2(\cosh^2 x - 1) - \cosh x - 1 = 0$ => $2\cosh^2 x - \cosh x - 3 = 0$. Factor: $(2\cosh x - 3)(\cosh x + 1) = 0$. Thus $\cosh x = \frac{3}{2}$ or $\cosh x = -1$. Since $\cosh x = \frac{3}{2}$, $x = \pm \cosh^{-1} \frac{3}{2} = \pm \ln \frac{3 + \sqrt{5}}{2} = \pm \ln(1 + \sqrt{2})$. The solution $\cosh x = -1$ is impossible since $\cosh x \ge 1$. This is a higher-order problem requiring substitution and solving a quadratic equation.

📖 Explanation: Using the formula \frac{d}{dx}[\cosh^{-1}u] = \frac{u'}{\sqrt{u^2 - 1}} with $u = 2x$, we get $\frac{2}{\sqrt{4x^2 - 1}}$. This requires knowing the derivative of the inverse hyperbolic cosine, which is often used in integration.

📖 Explanation: This integral is a standard form: $\int \frac{du}{\sqrt{u^2 - a^2}} = \cosh^{-1} \frac{u}{a} + C = \ln(u + \sqrt{u^2 - a^2}) + C$. Here $u = x$, $a = 3$, so the result is $\ln(x + \sqrt{x^2 - 9}) + C$. This is essential for finding arc lengths of certain curves.

📖 Explanation: Since $\cosh x = \frac{e^x + e^{-x}}{2} \ge 1$ for all real $x$, with equality only at $x = 0$. The minimum value is 1, and it grows without bound. This is crucial for understanding the shape of a hanging cable (catenary), where the lowest point is at $x = 0$.

📖 Explanation: $\cosh x = \sqrt{1 + \sinh^2 x} = \sqrt{1 + 9} = \sqrt{10}$. Then $\tanh x = \frac{\sinh x}{\cosh x} = \frac{3}{\sqrt{10}}$. This combines the fundamental identity with the definition of $\tanh$.

📖 Explanation: By definition, $\coth x = \frac{\cosh x}{\sinh x} = \frac{1}{\tanh x}$, $\operatorname{csch} x = \frac{1}{\sinh x}$, and $\operatorname{sech} x = \frac{1}{\cosh x}$. These are fundamental reciprocal relationships that define the hyperbolic functions. Understanding these is essential for simplifying complex expressions.

📖 Explanation: $\cosh x$ is an even function, so its graph is symmetric about the y-axis. This is because $\cosh(-x) = \cosh x$. The graph increases exponentially as $|x|$ increases. This symmetry is directly related to the shape of the catenary.

📖 Explanation: The first derivative is $\cosh x$. The second derivative is $\sinh x$. This means $\sinh x$ satisfies the differential equation y'' = y, which is a fundamental property used in solving many physical problems involving exponential growth or decay.

📖 Explanation: $\cosh x = \sqrt{1 + \sinh^2 x} = \sqrt{1 + 4} = \sqrt{5}$. Since $\cosh x$ is always positive, we take the positive root. This is a common operation when solving for a hyperbolic function from a given value.