Tag: Hyperbolic functions

Prove that the derivative of cosh x is sinh x

by RoRi

November 20, 2015

Prove the following formula for the derivative of hyperbolic cosine,

$D (\cosh x) = \sinh x.$

Proof. We can compute from the definition of hyperbolic cosine in terms of exponentials,

$\begin{align*} D(\cosh x) &= D \left( \frac{e^x + e^{-x}}{2} \right) \\ &= \frac{1}{2} (e^x - e^{-x}) \\ &= \sinh x. \qquad \blacksquare \end{align*}$

Prove that the derivative of sinh x is cosh x

by RoRi

November 20, 2015

Prove the following formula for the derivative of the hyperbolic sine,

$D (\sinh x) = \cosh x.$

Proof. We can compute the derivative from the definition of the hyperbolic sine in terms of exponentials,

$\begin{align*} D (\sinh x) &= D \left( \frac{e^x - e^{-x}}{2} \right) \\ &= \frac{1}{2} (e^x + e^{-x}) \\ &= \cosh x. \qquad \blacksquare \end{align*}$

Calculate tanh (2x) given tanh x = 3/4

by RoRi

November 19, 2015

Find the value of $\tanh(2x)$ given that $\tanh = \frac{3}{4}$ .

We recall the following formulas (from Section 6.19, Exercises #7 and #8),

$\begin{align*} \sinh(2x) &= 2 \sinh x \cosh x \\ \cosh (2x) &= \cosh^2 x + \sinh^2 x. \end{align*}$

Then using the definition of hyperbolic tangent we compute

$\begin{align*} \tanh (2x) &= \frac{\sinh (2x)}{\cosh (2x)} \\[9pt] &= \frac{2 \sinh x \cosh x}{\cosh^2 x + \sinh^2 x} \\[9pt] &= \frac{2 \operatorname{coth} x}{\operatorname{coth}^2 x + 1} \\[9pt] &= \frac{2 \left( \frac{4}{3} \right)}{\left( \frac{4}{3} \right)^2 + 1} \\[9pt] &= \frac{24}{25}. \end{align*}$

Calculate cosh (x+y) given sinh x = 4/3 and sinh y = 3/4

by RoRi

November 19, 2015

Find the value of $\cosh (x+y)$ given that $\sinh x = \frac{4}{3}$ and $\sinh y = \frac{3}{4}$ .

We use the identity $\cosh^2 = 1 + \sinh^2$ in both cases.

$\begin{align*} \cosh^2 x &= 1 + \sinh^2 x &\implies && \cosh^2 x &= 1 + \frac{16}{9} & \implies && \cosh x &= \frac{5}{3} \\ \cosh^2 y &= 1 + \sinh^2 y &\implies && \cosh^2 y &= 1 + \frac{9}{16} & \implies && \cosh y &= \frac{5}{4}. \end{align*}$

Then we recall (Section 6.19, Exercise #6) the formula for $\cosh (x+y)$ ,

$\cosh (x+y) = \cosh x \cosh y + \sinh x \sinh y.$

Therefore, we can compute

$\begin{align*} \cosh (x+y) &= \cosh x \cosh y + \sinh x \sinh y \\ &= \frac{5}{3} \cdot \frac{5}{4} + \frac{4}{3}\cdot \frac{3}{4} \\ &= \frac{25}{12} + 1 \\ &= \frac{37}{12}. \end{align*}$

Calculate sinh x and cosh x given tanh x = 5/13

by RoRi

November 19, 2015

Find the values of $\sinh x$ and $\cosh x$ given that $\tanh x = \frac{5}{13}$ .

Since $\tanh^2 + \sech^2 = 1$ (see here, Section 6.19 Exercise #14) we have

$\begin{align*} \tanh^2 x + \operatorname{sech}^2 x = 1 && \implies && \operatorname{sech}^2 x = \frac{144}{169} \\ && \implies && \operatorname{sech} x = \frac{12}{13} \\ && \implies && \cosh x = \frac{13}{12}. \end{align*}$

Then, to compute $\sinh x$ we have,

$\begin{align*} \tanh x = \frac{\sinh x}{\cosh x} && \implies && \frac{5}{13} &= \frac{12}{13} \sinh x \\ && \implies && \sinh x &= \frac{5}{12}. \end{align*}$

Calculate sinh x given cosh x = 5/4 and x > 0

by RoRi

November 19, 2015

Find the value of $\sinh x$ given that $\cosh x = \frac{5}{4}$ and $x > 0$ .

We know $\sinh^2 x = \cosh^2 x - 1$ . So, if $\cosh x = \frac{5}{4}$ we have

$\sinh^2 x = \frac{25}{16} - 1 = \frac{9}{16}.$

Thus, if $x > 0$ ,

$\sinh x = \frac{3}{4}.$

Calculate cosh x when sinh x = 4/3

by RoRi

November 19, 2015

Find the value of $\cosh x$ when given $\sinh x = \frac{4}{3}$ .

We know $\cosh^x x = 1 + \sinh^2 x$ . So, if $\sinh x = \frac{4}{3}$ then we have

$\cosh^2 x = 1 + \frac{16}{9} = \frac{25}{9}.$

Hence,

$\cosh x = \frac{5}{3}.$

Prove that coth² x – csch² x = 1

by RoRi

November 19, 2015

Prove the following identity,

$\operatorname{coth}^2 x - \operatorname{csch}^2 x = 1.$

Proof. From the definitions of hyperbolic cotangent and hyperbolic cosecant we have,

$\begin{align*} \operatorname{coth}^2 x - \operatorname{csch}^2 x &= \frac{\cosh^2 x}{\sinh^2 x} - \frac{1}{\sinh^2 x} \\ &= \frac{\cosh^2 x - 1}{\sinh^2 x} \\ &= \frac{\sinh^2 x}{\sinh^2 x} \\ &= 1. \qquad \blacksquare \end{align*}$

Prove that tanh² x + sech²x = 1

by RoRi

November 19, 2015

Prove the following identity,

$\tanh^2 x + \operatorname{sech}^2 x = 1.$

From the definitions of hyperbolic tangent and hyperbolic secant we have,

$\begin{align*} \tanh^2 x + \operatorname{sech}^2 x &= \frac{\sinh^2 x}{\cosh^2 x} + \frac{1}{\cosh^2 x} \\ &= \frac{\sinh^2 x + 1}{\cosh^2 x} \\ &= \frac{\cosh^2 x}{\cosh^2 x} \\ & = 1. \qquad \blacksquare \end{align*}$

Prove that 2 cosh² (x/2) = cosh x + 1

by RoRi

November 19, 2015

Prove the following identity,

$2 \cosh^2 \frac{1}{2}x = \cosh x + 1.$

Proof. Computing directly from the definition of the hyperbolic cosine,

$\begin{align*} 2 \cosh^2 \frac{x}{2} &= 2 \left( \frac{e^{\frac{x}{2}} + e^{-\frac{x}{2}}}{2} \right)^2 \\ &= 2 \left( \frac{e^x + 2 + e^{-x}}{4} \right) \\ &= \frac{e^x + e^{-x}}{2} + 1 \\ &= \cosh x + 1. \qquad \blacksquare \end{align*}$