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Hyperbolic Functions ET RIE versity ni o U f DEO PAT- Hyperbolic Functions ET RIE S 2 2 a n The trigonometric functions cos α and cos α are defined using the unit circle x +y = 1 sk a a ew by measuring the distance α in the counter-clockwise direction along the circumference tch α of the circle. The area of the sector so determined is , so we can equivalently say that 2 cos α and cos α are derived from the unit circle x2 + y2 = 1 by measuring off a sector α (shaded red )of area . The other four trigonometric functions can then be defined in 2 terms of cos and sin. Similarly, we may define hyperbolic functions cosh α and sinh α from the “unit hy- perbola” α x2 −y2 = 1 by measuring off a sector (shaded red )of area to obtain a point P whose 2 x- and y- coordinates are defined to be cosh α and sinh α. y y (cosh α, sinh α) x x Since at this point we do not yet know how to compute the areas of most curved regions, we must take it on faith that the six hyperbolic functions may be expressed simply in terms of the exponential function: eα − e−α eα + e−α sinh α = cosh α = 2 2 sinh α eα − e−α cosh α eα + e−α tanh α = = cotanh α = = cosh α eα + e−α sinh α eα − e−α 1 2 1 2 sech α = = cosech α = = cosh α eα + e−α sinh α eα − e−α Note that the domains of sinh, cosh, tanh, and sech are (−∞, ∞) and the domains of cotanh and cosech are (−∞, 0) ∪ (0, ∞). 1 eα + e−α eα − e−α We can check that the point , lies on the unit hyperbola: 2 2 eα + e−α 2 eα − e−α 2 e2α + 2 + e−2α e2α − 2 + e−2α 4 − = − = = 1 2 2 4 4 4 “Pythagorean” Identities This gives us the first important hyperbolic function identity: cosh2 α − sinh2 α ≡ 1 This may be used to derive two other identities relating the two other pairs of hyperbolic functions: 1 − tanh2 α = sech 2α and cotanh 2α − 1 = cosech 2α Odd and Even Identities It is clear that sinh, tanh, cotanh xand cosech are odd functions, while cosh, cotanh , and sech are even, so we have the corresponding identities: sinh(−x) =−sinh x, tanh(−x) =−tanh x, cotanh (−x) =−cotanh x, cosech (−x) =−cosech x cosh(−x) = cosh x, sech (−x) = sech x. Sum and Difference Identities We can use the above formulas for the hyperbolic functions in terms of ex to derive analogs of the identities for the trigonometric functions: eα − e−α eβ + e−β (eα − e−α)(eβ + e−β) sinh α cosh β = = = 2 2 4 eα+β + eα−β − e−α+β − e−α−β 4 eβ − e−β eα + e−α (eβ − e−β)(eα + e−α) sinh β cosh α = = = 2 2 4 2 eβ+α + eβ−α − e−β+α − e−β−α 4 Adding these two products gives: sinh α cosh β + sinh β cosh α = eα+β + eα−β − e−α+β − e−α−β eβ+α + eβ−α + e−β+α − e−β−α + = 4 4 2eα+β − 2e−α−β eα+β − e−α−β e(α+β) − e−(α+β) = = = sinh(α + β) 4 2 2 and subtracting these two products gives: sinh α cosh β − sinh β cosh α = eα+β + eα−β − e−α+β − e−α−β eβ+α + eβ−α + e−β+α − e−β−α − = 4 4 2eα−β − 2e−(α−β) eα−β − e−(α−β) = = sinh(α − β) 4 2 Similarly, eα + e−α eβ + e−β (eα + e−α)(eβ + e−β) cosh α cosh β = = = 2 2 4 eα+β + eα−β + eβ−α + e−α−β 4 eα − e−α eβ − e−β (eα − e−α)(eβ − e−β) sinh α sinh β = = = 2 2 4 eα+β − eα−β − eβ−α + e−α−β 4 Adding these two products gives cosh α cosh β + sinh α sinh β = eα+β + eα−β + eβ−α + e−α−β eα+β − eα−β − eβ−α + e−α−β + = 4 4 3 2eα+β + 2e−α−β eα+β + e−(α+β) = = cosh(α + β) 4 2 and subtracting them gives: cosh α cosh β − sinh α sinh β = eα+β + eα−β + eβ−α + e−α−β eα+β − eα−β − eβ−α + e−α−β − = 4 4 2eα−β + 2e−α+β eα−β + e−(α−β) = = cosh(α − β) 4 2 Summarizing, we have four identities: sinh(α + β) ≡ sinh α cosh β + sinh β cosh α sinh(α − β) ≡ sinh α cosh β − sinh β cosh α cosh(α + β) ≡ cosh α cosh β + sinh α sinh β cosh(α − β) ≡ cosh α cosh β − sinh α sinh β which are almost exactly parallel to those for the trigonometric functions and may be used to derive sum and difference formulas for the other four hyperbolic functions. Double and Half-“Angle” Identities Letting β = α, we get: sinh 2α ≡ 2 sinh α cosh α, cosh 2α ≡ cosh2 α + sinh2 α ≡ 1 + 2 sinh2 α ≡ 2 cosh2 α − 1, so cosh 2α + 1 cosh 2α − 1 cosh2 α = and sinh2 α = , and thus: 2 2 cosh 2α + 1 cosh 2α − 1 cosh α = and sinh α = 2 2 4 α cosh α + 1 α cosh α − 1 cosh = and sinh = 2 2 2 2 Derivatives d d ex − e−x ex − (−e−x) ex + e−x (sinh x) = = = = cosh x dx dx 2 2 2 d d ex + e−x ex + (−e−x) ex − e−x (cosh x) = = = = sinh x dx dx 2 2 2 d d sinh x (tanh x) = = dx dx cosh x cosh x(sinh x) − sinh x(cosh x) cosh x cosh x − sinh x sinh x = = cosh2 x cosh2 x cosh2 x − sinh2 x 1 = = sech 2x cosh2 x cosh2 x d d cosh x (cotanh x) = = dx dx sinh x sinh x(cosh x) − cosh x(sinh x) sinh x sinh x − cosh x cosh x = = sinh2 x sinh2 x sinh2 x − cosh2 x −1 = =−cosech 2x sinh2 x sinh2 x d d (sech x) = (cosh x)−1 = dx dx (−1) (cosh x)−2 (cosh x) = (−1) (cosh x)−2 sinh x =−sech x tanh x d d (cosech x) = (sinh x)−1 = dx dx 5 (−1) (sinh x)−2 (sinh x) = (−1) (sinh x)−2 cosh x =−cosech xcotanh x Summary: d d (sinh x) = cosh x (cosh x) = sinh x dx dx d d (tanh x) = sech 2x (cotanh x) =−cosech 2x dx dx d d (sech x) =−sech x tanh x (cosech x) =−cosech xcotanh x dx dx Graphs ofthe Hyperbolic Functions y y y x x x y = sinh x y = cosh x y = tanh x y = cosech x y = sech x y = cotanh x The domains and ranges are summarized in the next table: 6 function domain Range sinh (−∞, ∞) (−∞, ∞) cosh (−∞, ∞) [1, ∞) tanh (−∞, ∞) (−1, 1) cotanh (−∞, 0) ∪ (0, ∞) (−∞, −1) ∪ (1, ∞) sech (−∞, ∞) (0, 1]) cosech (−∞, 0) ∪ (0, ∞) (−∞, 0) ∪ (0, ∞) Inverse Hyperbolic Functions sinh, tanh, cotanh and cosech are one-to-one, but cosh and sech are not. For the purpose of defining the inverse of cosh and sech we will restrict their domains to [0, ∞). We will denote the inverse hyperbolic functions by sinh−1, cosh−1, tanh−1, cotanh −1, sech −1, and cosech −1 or: sinhinv , coshinv , tanhinv , cotanh inv , sech inv , and cosech inv or even: arcsinh , arccosch , arctanh , arccothh , arcsech , and arccosech . The usual Cancellation Laws hold in the appropriate domains: sinh(sinh−1 x) ≡ x sinh−1(sinh x) ≡ x cosh(cosh−1 x) ≡ x cosh−1(cosh x) ≡ x tanh(tanh−1 x) ≡ x tanh−1(tanh x) ≡ x cotanh (cotanh −1x) ≡ x cotanh −1(cotanh x) ≡ x sech (sech −1x) ≡ x sech −1(sech x) ≡ x cosech (cosech −1x) ≡ x cosech −1(cosech x) ≡ x The derivatives of the inverse hyperbolic functions may be found the same way the derivatives of the inverse trigonometric functions were found: by differentiating the left-hand Cancellation Laws above: Example: Differentiating sinh(sinh−1 x) ≡ x we get 7 cosh(sinh−1 x) sinh−1 x = 1, so 1 sinh−1 x = . cosh(sinh−1 x) Using the identity cosh2 x − sinh2 x ≡ 1weget cosh2 x ≡ 1 + sinh2 x,so cosh x ≡ 1 + sinh2 x and therefore 2 cosh(sinh−1 x) = 1 + sinh2(sinh−1 x) = 1 + sinh(sinh−1 x) = 1 + x2 1 Thus we have sinh−1 x = √ 1 + x2 One may similarly derive the derivatives of the other hyperbolic functions: 1 cosh−1 x = √ x − 12 1 tanh−1 x = cotanh −1x = 1 − x2 1 sech −1x =− √ x 1 − x2 −1 cosech −1x = √ |x| 1 + x2 Explicit Computation ofInverse Hyperbolics The inverse hyperbolic functions have the unusual property that they can be explicity computed: Example: Solve the equation sinh y = x for y in terms of x. (The solution will be sinh−1 x!) ey − e−y We have sinh y = = x,so 2 ey −e−y = 2x or ey −2x −e−y = 0. Multiplying both sides of this equation by ey we get: 8 (ey )2 − 2xey − 1 = 0, a quadratic equation in ey which has solution √ −(−2x) ± (−2x)2 − 4(1)(−1) 2x ± 4x2 + 4 ey = = = x ± x2 + 1 2 2 √ Since x − x2 + 1 < 0 and we must have ey > 0, we get ey = x + x2 + 1. Taking logarithms of both sides of this equation, we get √ y = ln(x + x2 + 1), so we have sinh−1 x = ln(x + x2 + 1) Similarly, 1 1 + x cosh−1 x = ln(x + x2 − 1) and tanh−1 x = ln 2 1 − x We then have 1 1 1 2 1 1 + x2 cosech −1x = sinh−1 = ln + + 1 = ln + = x x x x x2 √ 1 1 + x2 ln + x |x| Similarly √ 1 x + 1 1 + 1 − x2 cotanh −1x = ln and sech −1x = ln 2 x − 1 x 9.
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