Write down the definitions of \( \sinh x \) and \( \cosh x \) in terms of exponentials.
\( \sinh x = \dfrac{e^x - e^{-x}}{2} \), \( \cosh x = \dfrac{e^x + e^{-x}}{2} \).
Hyperbolic Functions
12 exam-style questions on definitions, identities, inverse hyperbolics, differentiation and integration. Tap Show answer for full working.
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Prove that \( \cosh^2 x - \sinh^2 x = 1 \).
\( \cosh^2 x - \sinh^2 x = \Big(\dfrac{e^x+e^{-x}}{2}\Big)^2 - \Big(\dfrac{e^x-e^{-x}}{2}\Big)^2 = \dfrac{1}{4}\big[(e^{2x}+2+e^{-2x}) - (e^{2x}-2+e^{-2x})\big] = \dfrac{1}{4}\cdot 4 = 1 \).
Differentiate \( \sinh x \) and \( \cosh x \) from first principles using the exponential definitions.
\( \dfrac{d}{dx}\sinh x = \dfrac{e^x + e^{-x}}{2} = \cosh x \).
\( \dfrac{d}{dx}\cosh x = \dfrac{e^x - e^{-x}}{2} = \sinh x \).
Solve \( \sinh x = 0 \).
\( \dfrac{e^x - e^{-x}}{2} = 0 \) ⇒ \( e^x = e^{-x} \) ⇒ \( e^{2x} = 1 \) ⇒ \( x = 0 \).
Express \( \tanh x \) in terms of exponentials and find \( \lim_{x\to\infty} \tanh x \).
\( \tanh x = \dfrac{\sinh x}{\cosh x} = \dfrac{e^x - e^{-x}}{e^x + e^{-x}} = \dfrac{e^{2x} - 1}{e^{2x} + 1} \).
As \( x\to\infty \), \( \tanh x \to 1 \).
Prove the identity \( \sinh 2x = 2\sinh x\cosh x \).
RHS: \( 2\cdot\dfrac{e^x-e^{-x}}{2}\cdot\dfrac{e^x+e^{-x}}{2} = \dfrac{(e^x-e^{-x})(e^x+e^{-x})}{2} = \dfrac{e^{2x} - e^{-2x}}{2} = \sinh 2x \).
Find \( \dfrac{d}{dx}(\sinh^{-1}x) \) and hence evaluate \( \displaystyle\int \dfrac{1}{\sqrt{x^2+1}}\, dx \).
Let \( y = \sinh^{-1}x \), so \( \sinh y = x \), \( \cosh y\cdot y' = 1 \). Since \( \cosh y = \sqrt{1+\sinh^2 y} = \sqrt{1+x^2} \), \( y' = \dfrac{1}{\sqrt{1+x^2}} \).
Thus \( \displaystyle\int \dfrac{1}{\sqrt{x^2+1}}\,dx = \sinh^{-1}x + C \) (or \( \ln(x + \sqrt{x^2+1}) + C \)).
Solve \( 2\cosh x + \sinh x = 2 \).
Write in exponentials: \( 2\cdot\dfrac{e^x+e^{-x}}{2} + \dfrac{e^x-e^{-x}}{2} = 2 \) ⇒ \( e^x + e^{-x} + \dfrac12 e^x - \dfrac12 e^{-x} = 2 \).
\( \dfrac32 e^x + \dfrac12 e^{-x} = 2 \). Multiply by \( 2e^x \): \( 3e^{2x} + 1 = 4e^x \) ⇒ \( 3e^{2x} - 4e^x + 1 = 0 \).
\( (3e^x - 1)(e^x - 1) = 0 \) ⇒ \( e^x = 1 \) or \( e^x = 1/3 \) ⇒ \( x = 0 \) or \( x = -\ln 3 \).
Find the exact value of \( \cosh^{-1} 2 \) in logarithmic form.
\( \cosh y = 2 \) ⇒ \( \dfrac{e^y + e^{-y}}{2} = 2 \) ⇒ \( e^y + e^{-y} = 4 \). Multiply by \( e^y \): \( e^{2y} - 4e^y + 1 = 0 \).
\( e^y = 2 \pm \sqrt{3} \). Since \( y > 0 \) for principal value, \( e^y = 2 + \sqrt{3} \).
\( \cosh^{-1} 2 = \ln(2 + \sqrt{3}) \).
Differentiate \( y = \tanh^{-1}x \) and show that \( \dfrac{dy}{dx} = \dfrac{1}{1-x^2} \) for \( |x| < 1 \).
\( \tanh y = x \), \( \operatorname{sech}^2 y \cdot y' = 1 \). But \( \operatorname{sech}^2 y = 1 - \tanh^2 y = 1 - x^2 \).
Hence \( y' = \dfrac{1}{1-x^2} \).
Evaluate \( \displaystyle\int_0^{\ln 2} \sinh^2 x\, dx \).
\( \sinh^2 x = \dfrac{\cosh 2x - 1}{2} \).
\( \displaystyle\int_0^{\ln 2} \dfrac{\cosh 2x - 1}{2}\,dx = \dfrac12\Big[\dfrac12\sinh 2x - x\Big]_0^{\ln 2} = \dfrac14\sinh(2\ln 2) - \dfrac12\ln 2 \).
\( \sinh(2\ln 2) = \dfrac{e^{2\ln 2} - e^{-2\ln 2}}{2} = \dfrac{4 - 1/4}{2} = \dfrac{15}{8} \).
Result: \( \dfrac{15}{32} - \dfrac12\ln 2 \).
Solve the equation \( 3\sinh x - 4\cosh x = -4 \) for real \( x \).
Write \( 3\dfrac{e^x-e^{-x}}{2} - 4\dfrac{e^x+e^{-x}}{2} = -4 \) ⇒ \( \dfrac{3}{2}e^x - \dfrac{3}{2}e^{-x} - 2e^x - 2e^{-x} = -4 \).
\( (-\dfrac12)e^x - \dfrac72 e^{-x} = -4 \). Multiply by \( -2e^x \): \( e^{2x} + 7 = 8e^x \) ⇒ \( e^{2x} - 8e^x + 7 = 0 \).
\( (e^x - 1)(e^x - 7) = 0 \) ⇒ \( x = 0 \) or \( x = \ln 7 \).
Check both satisfy the original (both do).