Hyperbolic Trigonometric Functions and Their Identities - Prof. William Faucette, Study notes of Calculus

Download Hyperbolic Trigonometric Functions and Their Identities - Prof. William Faucette and more Study notes Calculus in PDF only on Docsity! Hyperbolic Trigonometric Functions Just as trigonometry can be performed on the unit circle, it can also be performed on the unit hyperbola: x2 − y2 = 1. Define the hyperbolic cosine function by cosh(x) = ex + e−x 2 and the hyperbolic sine function by sinh(x) = ex − e−x 2 . A simple computation then shows that cosh2(x) − sinh2(x) = 1, which is the hyperbolic trigonometric identity analogous to the Pythagorean identity cos2(x) + sin2(x) = 1 from ordinary (circular) trigonometry. Next, as in the case of ordinary trigonometry, we define the remaining four hyperbolic trigonometric functions from these first two: Define the hyperbolic tangent function by tanh(x) = sinh(x) cosh(x) and the hyperbolic cotangent function by coth(x) = cosh(x) sinh(x) . Define the hyperbolic secant function by sech(x) = 1 cosh(x) and the hyperbolic cosecant function by csch(x) = 1 sinh(x) . Hyperbolic Pythagorean Identities In addition to the identity derived earlier, cosh2(x) − sinh2(x) = 1, there are two other hyperbolic identities obtained by dividing this one by sinh2(x) and cosh2(x), respectively: cosh2(x) − sinh2(x) = 1 cosh2(x) sinh2(x) − sinh2(x) sinh2(x) = 1 sinh2(x) ( cosh(x) sinh(x) )2 − 1 = 1 sinh2(x) coth2(x) − 1 = csch2(x). 1 cosh2(x) − sinh2(x) = 1 cosh2(x) cosh2(x) − sinh2(x) cosh2(x) = 1 cosh2(x) 1 − ( sinh(x) cosh(x) )2 = 1 cosh2(x) 1 − tanh2(x) = sech2(x). We summarize these here: cosh2(x) − sinh2(x) = 1 1 − tanh2(x) = sech2(x) coth2(x) − 1 = csch2(x) Derivatives of Hyperbolic Functions By using the definition of the hyperbolic sine function, we have d dx (sinh(x)) = d dx ( ex − e−x 2 ) = ex + e−x 2 = cosh(x). Similarly, d dx (cosh(x)) = d dx ( ex + e−x 2 ) = ex − e−x 2 = sinh(x). The derivatives of the remaining hyperbolic trigonometric functions are obtained from their definitions using the Pythagorean hyperbolic trigonometric identities, the definitions of the hyperbolic functions, and the quotient rule: d dx (tanh(x)) = d dx ( sinh(x) cosh(x) ) = d dx (sinh(x)) (cosh(x)) − (sinh(x)) d dx (cosh(x)) cosh2(x) = cosh(x) cosh(x) − sinh(x) sinh(x) cosh2(x) = cosh2(x) − sinh2(x) cosh2(x) = 1 cosh2(x) = sech2(x). Similarly, d dx (coth(x)) = d dx ( cosh(x) sinh(x) ) = d dx (cosh(x)) (sinh(x)) − (cosh(x)) d dx (sinh(x)) sinh2(x) = sinh(x) sinh(x) − cosh(x) cosh(x) sinh2(x) = sinh2(x) − cosh2(x) sinh2(x) = −1 sinh2(x) = − csch2(x). 2 so x = ey − e−y ey + e−y . Clearing fractions and moving everything to one side, we get x ( ey + e−y ) = ey − e−y x ( e2y + 1 ) = e2y − 1 xe2y + x = e2y − 1 1 + x = e2y − xe2y 1 + x = e2y (1 − x) e2y = 1 + x 1 − x 2y = ln ( 1 + x 1 − x ) y = 1 2 ln ( 1 + x 1 − x ) So, tanh−1(x) = 1 2 ln ( 1 + x 1 − x ) Logarithmic form for coth−1(x) Let y = coth−1(x). Then x = coth(y), so x = ey + e−y ey − e−y . Clearing fractions and moving everything to one side, we get x ( ey − e−y ) = ey + e−y x ( e2y − 1 ) = e2y + 1 xe2y − x = e2y + 1 x + 1 = xe2y − e2y x + 1 = e2y (x − 1) e2y = x + 1 x − 1 2y = ln ( x + 1 x − 1 ) y = 1 2 ln ( x + 1 x − 1 ) So, coth−1(x) = 1 2 ln ( 1 + x x − 1 ) 5 Logarithmic form for sech−1(x) Let y = sech−1(x). Then x = sech(y), so x = 1 cosh(y) = 2 ey + e−y . Clearing fractions and moving everything to one side, we get x ( ey + e−y ) − 2 = 0 x ( e2y + 1 ) − 2ey = 0 xe2y − 2ey + x = 0. This is a quadratic equation in ey , so we can solve this using the quadratic formula: ey = 2 ± √ (−2)2 − 4(x)(x) 2x = 2 ± √ (−2)2 − 4(x)(x) 2x = 2 ± √ 4 − 4x2 2x = 2 ± 2 √ 1 − x2 2x = 2 ( 1 ± √ 1 − x2 ) 2x = 1 ± √ 1 − x2 x Since ey must be positive, we must have ey = 1 + √ 1 − x2 x . Taking the natural logarithm, we get sech−1(x) = ln ( 1 + √ 1 − x2 x ) . Logarithmic form for csch−1(x) Let y = csch−1(x). Then x = csch(y), so x = 1 sinh(y) = 2 ey − e−y . Clearing fractions and moving everything to one side, we get x ( ey − e−y ) − 2 = 0 x ( e2y − 1 ) − 2ey = 0 xe2y − 2ey − x = 0. 6 This is a quadratic equation in ey , so we can solve this using the quadratic formula: ey = 2 ± √ (−2)2 − 4(x)(−x) 2x = 2 ± √ 4 + 4x2 2x = 2 ± 2 √ 1 + x2 2x = 2 ( 1 ± √ 1 + x2 ) 2x = 1 ± √ 1 + x2 x Since ey must be positive, we must have ey =        1 − √ 1 + x2 x for x < 0 1 + √ 1 + x2 x for x > 0 We can write these two equations as one as follows: ey = 1 x + √ 1 + x2 |x | Taking the natural logarithm, we get csch−1(x) = ln ( 1 x + √ 1 + x2 |x | ) . Derivatives of Inverse Hyperbolic Functions Verify the following derivatives: d dx ( sinh−1(x) ) = 1 √ 1 + x2 d dx ( cosh−1(x) ) = 1 √ x2 − 1 d dx ( tanh−1(x) ) = 1 1 − x2 d dx ( coth−1(x) ) = 1 1 − x2 d dx ( sech−1(x) ) = − 1 x √ 1 − x2 d dx ( csch−1(x) ) = − 1 |x | √ 1 + x2 7