Hyperbolic Functions Some combinations of ex and e-x arise in solving some differential equations in physics and Engineering. These are called Hyperbolic Functions Some Applications : Motion of waves in Elastic Solids, the Shape of Electric Power lines, Persuit Curves, Geometry of the general theory of relativity etc. These functions are related to the exponential and circular functions. def : hyperbolic cosine HcoshL : hyperbolic sine HsinhL : 1 sinh x = H 2 1 cosh x = H 2 ex + e-x L ex - e-x L x2 x4 x6 cosh x = 1 + + + + ... 2! 4! 6! x3 x5 x7 sinh = x + + + + ... 3! 5! 7! For each real t cos2 t + sin2 t = 1 thus the point Hcos t, sin t L lies on the unit circle : x2 + y2 = 1. Hence sin and cos are called "circular functions " Similarly for real t cohs2 t - sinh2 t = 1 thus the point H cosh t, sinh tL lies on the hyperbola : x2 - y2 = 1. Hence sinh H cinchL and cosh HkoshL are called "hyperbolic functions" Other hyperbolic functions : Analogous to circular functionsM : sinh x H ex + e-x L tanh x = = cosh x H ex - e-x L 1 sech x = cosh x 1 cosech x = sinh x 1 coth x = tanh x Graphs of cosh x and sinh x 1 cosh x = H 2 ex + e-x L , 1 sinh x = H 2 1 2 sech x = = cosh x Hex + e-x L 1 cosh 0 = H 2 1 e0 + e-0 L = H1 + 1L = 1 2 1 2 cosech x = = sinh x H ex - e-x L 1 sinh 0 = H 2 1 e0 - e-0 L = H1 - 1L = 0 2 1 ex + e-x coth x = = tanh x ex - e-x ex - e-x L Inverse Hyperbolic Functions : HAnalogous to sin-1 x and cos-1 xL Logarithmic Forms : y = sinh-1 x means x = sinh y or multiply by ey ey - e-y L or 2 x = ey - e-y and rearrange to e2 y - 2 x ey - 1 = 0 ey = u, so Set 1 x = H 2 e2 y = u2 , to get u2 - 2 xu - 1 = 0 Hquadratic in uL !!!!!!!!!!!!!!!!! 2 x + - 4 x2 + 4 u = ey = 2 !!!!!!!!!!!!! = x + - x2 + 1 Roots : so ey = x + - sign is rejected since ey > 0 !!!!!!!! !!!!! x2 + 1 Taking ln and noting ln e = 1 y = sinh-1 x = ln Ax + !!!!!!!! !!!!! x2 + 1 E for all x. cosh-1 x = ln Ax + !!!!!!!! !!!!! x2 - 1 E x³1 Similarly Example : y = tanh-1 x ey - e-y e2 y - 1 x = tanh y = = , ey + e-y e2 y + 1 hence so x He2 y + 1L = e2 y - 1, 1+x e2 y = 1-x Since e2 y > 0 so - 1 < x < 1 or È x È < 1. Taking ln we get 1+x 2 y ln e = ln 1-x or 1 1+x y = tanh-1 x = ln , 2 1-x ÈxÈ <1 Hyperbolic Identities : 1 cosh x = Hex + e-x L 2 From and we get and 1 sinh x = Hex - e-x L 2 cosh x + sinh x = ex cosh x - sinh x = e-x Multiply last two equations to get cosh 2 x - sinh2 x = 1 H cos 2 x + sin2 x = 1L From this we can get others sech 2 x = 1 - tanh2 x H sec 2 x = 1 + tan2 xL cosech 2 x = coth2 x - 1 Hcosec 2 x = 1 + cot2 xL Now from definition; 1 cosh Hx + yL = Hex+y + e-Hx+yL L 2 1 = @ex ey + e-x e-y D 2 1 = @ 2 or Hcosh x + sinh xL Hcosh y + sinh yL + Hcosh x - sinh xL Hcosh y - sinh yL cosh Hx + yL = cosh x cosh y + sinh x sinh y Hcos Hx + yL = cos x cos y - sin x sin yL Setting y = x gives cosh 2 x = cosh2 x + sinh2 x H cos 2 x = cos2 x - sin2 xL D From this it is easy to show : cosh 2 x = 2 cosh2 x - 1 = 1 + 2 sinh2 x Hcos 2 x = 2 cos2 x - 1 = 1 - 2 sin2 xL sinh Hx + yL = sinh x cosh y + cosh x sinh y Hsin Hx + yL = sin x cos y + cos x sin yL Similarly sinh 2 x = 2 sinh x cosh x Hsin 2 x = 2 sin x cos xL 2 tanh x tanh 2 x = 1 + tanh2 x And 2 tan x tan 2 x = 1 - tan2 x etc Recall Euler’ s Relation ejΘ = cos Θ + jsin Θ e-jΘ = cos Θ - jsin Θ By adding and subtracting 1 cos Θ = H ejΘ + e-jΘ L 2 1 sin Θ = H ejΘ - e-jΘ L 2j Now 1 cosh x = H ex + e-x L 2 for x = jΘ one gets 1 cosh jΘ = H ejΘ + e-jΘ L = cos Θ 2 or Similarly gives Hence cosh jΘ = cos Θ ... ... .. HiL 1 sinh x = H ex - e-x L 2 1 sinh jΘ = H ejΘ - e-jΘ L = j sin Θ 2 sinh jΘ = j sin Θ ... ... .. HiiL Put Θ = jx in HiL to get cos jx = cosh j2 x = cosh H-xL = cosh x or cos jx = cosh x ... ... ... ....HiiiL Put Θ = jx in HiiL to get j sin jx = sinh j2 x = sinh H-xL = -sinh x Hence or j sin jx = j2 sinh x sin jx = j sinh x ... ... ... ... .. HivL Important Results : sin jx = j sinh x sinh jx = j sin x cos jx = cosh x cosh jx = cos x tan jx = j tanh x tanh jx = j tan x Example : 5 solve sin z = - 4 Let z = x + jy, then NB. z cannot be real sin Hx + jyL = sin x cos jy + cos x sin jy = sin x cosh y + j cos x sinh y Hence sin z = sin Hx + jyL = sin x cosh y + j cos x sinh y 5 = - 4 Thus 5 sin x cosh y = - ... ... ....HiL 4 cos x sinh y = 0. ... ... ... ....HiiL From HiiL either sinh y = 0 ie y = 0, or Π 3Π x = +- , +- , ... 2 2 cos x = 0, HaL If y = 0 then cosh y = 1 and from HiL Do you believe that ? HbL If cos x = 0, then sin x = +1 or - 1 HiL sin x = +1, then from HiL HiiL sin x = -1, ie. Also now from HiL 5 sin x = - 4 Hwhy?L 5 cosh y = - 4 5Π Π 3Π 7Π x =. .. - , - , , , ... 2 2 2 2 5 cosh y = 4 or or 1 5 Hey + e-y L = , 2 4 2 ey + 2 e-y - 5 = 0 2 e2 y - 5 ey + 2 = 0 or A quadratic in ey with factors H2 ey - 1L Hey - 2L = 0 1 Roots : ey = and ey = 2 2 Taking ln : 1y j y = ln i z or j z k2{ and y = ln 2. Hence z = x + yj HImpossibleL y = -ln 2 Π 3Π 7Π x =. .., - , , , ... 2 2 2 y = +-ln 2