H6, H7 - Hyperbolic identities

H6, H7 - Hyperbolic identities

The point (cosh⁡t,sinh⁡t)(\cosh t,\sinh t) moves along a hyperbola, which explains the name “hyperbolic”. You will derive the identities behind that connection, use double-angle formulae, and build proofs from exponential definitions with the necessary domain restrictions.

Ratios and squared identities

Why does the pair (5/4,3/4)(5/4,3/4) fit a hyperbola rather than the unit circle? Its squared difference is 25/16−9/16=125/16-9/16=1, while its squared sum is 34/1634/16, rather than 1.

This numerical difference of squares has a general counterpart. Starting from the exponential definitions gives

cosh⁡2x−sinh⁡2x=(ex+e−x)2−(ex−e−x)24=4exe−x4=1\begin{aligned} \cosh^2x-\sinh^2x &=\frac{(e^x+e^{-x})^2-(e^x-e^{-x})^2}{4}\\ &=\frac{4e^xe^{-x}}4=1 \end{aligned}

Consequently, setting X=cosh⁡tX=\cosh t and Y=sinh⁡tY=\sinh t gives X2−Y2=1X^2-Y^2=1, with X≥1X\geq1: a point on the right branch of a hyperbola. The minus sign is the key difference from the unit-circle identity.

Since cosh is never zero, divide the identity by cosh⁡2x\cosh^2x. For the second reciprocal identity divide by sinh⁡2x\sinh^2x, which requires x≠0x\ne0.

Squared hyperbolic identities

tanh⁡x=sinh⁡xcosh⁡x,cosh⁡2x−sinh⁡2x=1\tanh x=\frac{\sinh x}{\cosh x},\qquad \cosh^2x-\sinh^2x=1 sech⁡2x=1−tanh⁡2x(x∈R)cosech⁡2x=coth⁡2x−1(x≠0)\operatorname{sech}^2x=1-\tanh^2x\quad(x\in\mathbb R) \qquad \operatorname{cosech}^2x=\coth^2x-1\quad(x\ne0)

The fundamental identity cosh⁡2x−sinh⁡2x=1\cosh^2x-\sinh^2x=1 is in the formula booklet. The tanh quotient and the two reciprocal-square identities are not: learn or derive them.

An identity determines a square; a sign still comes from the range or the given input condition. Cosh is always positive, whereas sinh has the same sign as its input.

Worked example 1: a positive input

Question. Given sinh⁡t=3/4\sinh t=3/4, find cosh⁡t\cosh t and tanh⁡t\tanh t.

Plan. Obtain the cosh square from the fundamental identity, choose its positive root, then divide.

cosh⁡2t=1+916=2516,cosh⁡t=54,tanh⁡t=3/45/4=35\cosh^2t=1+\frac9{16}=\frac{25}{16},\qquad \cosh t=\frac54,\qquad \tanh t=\frac{3/4}{5/4}=\frac35

Answer. cosh⁡t=5/4\cosh t=5/4, tanh⁡t=3/5\tanh t=3/5.

Check. The reciprocal identity gives sech⁡2t+tanh⁡2t=16/25+9/25=1\operatorname{sech}^2t+\tanh^2t=16/25+9/25=1.

Change only the sign of the given sinh value. The positive cosh choice remains essential.

Worked example 2: a negative input

Question. Given sinh⁡t=−3/4\sinh t=-3/4, find cosh⁡t\cosh t and tanh⁡t\tanh t.

Plan. Use the same squared identity and choose cosh from its range.

cosh⁡2t=1+(−34)2=2516,cosh⁡t=54,tanh⁡t=−35\cosh^2t=1+\left(-\frac34\right)^2=\frac{25}{16},\qquad \cosh t=\frac54,\qquad \tanh t=-\frac35

Answer. cosh⁡t=5/4\cosh t=5/4, tanh⁡t=−3/5\tanh t=-3/5.

Check. Here coth⁡t=−5/3\coth t=-5/3 and cosech⁡t=−4/3\operatorname{cosech}t=-4/3; their squares satisfy 25/9−16/9=125/9-16/9=1.

Use the sign of the input when the unknown is sinh rather than cosh.

Watch out

Wrong: cosh⁡t=±5/4\cosh t=\pm5/4 after finding cosh⁡2t=25/16\cosh^2t=25/16. Cosh is always positive, so its value is 5/45/4. An unknown sinh value instead needs a sign chosen from the input.

Here the input condition fixes that sign for you.

Double-angle identities

Doubling the input squares the exponentials. Expanding the definitions shows

2sinh⁡xcosh⁡x=(ex−e−x)(ex+e−x)2=e2x−e−2x2=sinh⁡2xcosh⁡2x+sinh⁡2x=(ex+e−x)2+(ex−e−x)24=e2x+e−2x2=cosh⁡2x\begin{aligned} 2\sinh x\cosh x &=\frac{(e^x-e^{-x})(e^x+e^{-x})}{2} =\frac{e^{2x}-e^{-2x}}2=\sinh2x\\ \cosh^2x+\sinh^2x &=\frac{(e^x+e^{-x})^2+(e^x-e^{-x})^2}{4} =\frac{e^{2x}+e^{-2x}}2=\cosh2x \end{aligned}

These identities hold for every real input. Combining the second with cosh⁡2x−sinh⁡2x=1\cosh^2x-\sinh^2x=1 gives three useful versions:

Double angles and squares

sinh⁡2x=2sinh⁡xcosh⁡xcosh⁡2x=cosh⁡2x+sinh⁡2x=2cosh⁡2x−1=1+2sinh⁡2x\sinh2x=2\sinh x\cosh x \qquad \cosh2x=\cosh^2x+\sinh^2x=2\cosh^2x-1=1+2\sinh^2x

Thus cosh⁡2x=(cosh⁡2x+1)/2\cosh^2x=(\cosh2x+1)/2 and sinh⁡2x=(cosh⁡2x−1)/2\sinh^2x=(\cosh2x-1)/2.

The two double-angle identities are in the formula booklet; the square forms follow by rearranging them with the fundamental identity.

The two square formulae differ only by the sign of the constant. They will turn squares into functions that integrate directly in the next lesson.

Worked example 3: doubling an input

Question. Given sinh⁡t=3/4\sinh t=3/4 and cosh⁡t=5/4\cosh t=5/4, find sinh⁡2t\sinh2t and cosh⁡2t\cosh2t.

Plan. Substitute in the two double-angle identities.

sinh⁡2t=2(34)(54)=158,cosh⁡2t=2516+916=178\sinh2t=2\left(\frac34\right)\left(\frac54\right)=\frac{15}{8},\qquad \cosh2t=\frac{25}{16}+\frac9{16}=\frac{17}{8}

Answer. sinh⁡2t=15/8\sinh2t=15/8, cosh⁡2t=17/8\cosh2t=17/8.

Check. Their squared difference is (289−225)/64=1(289-225)/64=1, and the cosh value is at least 1.

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