Tag: Identities

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,

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

Prove that 2 sinh² (x/2) = cosh x – 1

by RoRi

November 19, 2015

Prove the following identity:

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

Proof. We compute directly from the definition of $\sinh x$ ,

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

Prove that (cosh x + sinh x)ⁿ = cosh (nx) + sinh (nx)

by RoRi

November 18, 2015

Prove the following identity:

$(\cosh x + \sinh x)^n = \cosh (nx) + \sinh (nx).$

Proof. We know from a previous exercise (Section 6.19, Exercise #9) that

$\cosh x + \sinh x = e^x \quad \implies \quad \cosh (nx) + \sinh(nx) = e^{nx}.$

Therefore, we have

$(\cosh x + \sinh x)^n = (e^x)^n = e^{nx} = \cosh (nx) + \sinh(nx). \qquad \blacksquare$

Prove that cosh x – sinh x = e^-x

by RoRi

November 18, 2015

Prove that

$\cosh x - \sin x = e^{-x}.$

Proof. Using the definition of the hyperbolic functions in terms of the exponential, we have

$\begin{align*} \cosh x - \sinh x &= \frac{e^x+e^{-x}}{2} - \frac{e^x - e^{-x}}{2} \\ &= \frac{e^x + e^{-x} - e^x + e^{-x}}{2} \\ &= \frac{2e^{-x}}{2} \\ &= e^{-x}. \qquad \blacksquare \end{align*}$

Prove that cosh x + sinh x = e^x

by RoRi

November 18, 2015

Prove that

$\cosh x + \sinh x = e^x.$

Proof. We use the definitions of hyperoblic sine and cosine to compute,

$\begin{align*} \cosh x + \sinh x &= \frac{e^x+e^{-x}}{2} + \frac{e^x - e^{-x}}{2} \\ &= \frac{e^x + e^{-x} + e^x -e^{-x}}{2} \\ &= \frac{2e^x}{2} \\ & = e^x. \qquad \blacksquare \end{align*}$

Prove that cosh (2x) = cosh²x + sinh² x

by RoRi

November 17, 2015

Prove that

$\cosh (2x) = \cosh^2 x + \sinh^2 x.$

Proof. We know from this previous exercise (Section 6.19, Exercise #6) that

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

Therefore,

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

Prove that sinh(2x) = 2 sinh x cosh x

by RoRi

November 17, 2015

Prove that

$\sinh (2x) = 2 \sinh x \cosh x.$

Proof. We know from this exercise (Section 6.19, Exercise #5) that

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

Therefore,

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

Prove a formula for cosh (x+y)

by RoRi

November 17, 2015

Prove that

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

Proof. Using the definition of the hyperbolic cosine in terms of the exponential we have,

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