Hyperbolic Functions Notice: this material must not be use as a substitute for attening the lectures
0. Hyperbolic functions sinh an cosh The hyperbolic functions sinh (pronounce shine ) an cosh are efine by the formulae cosh x ex + e x sinh x ex e x () The function cosh x is an even function, an sinh x is o. On moern calculators hyperbolic functions are usually accesse using a button marke hyp. So the sinh function woul be accesse by typically using a sequence of keystrokes of the form hyp sin. The function y cosh x (or more precisely y a cosh x/a for a suitable value of a) gives the exact shape of a heavy chain or rope hanging between two fixe points. The cosh an sinh functions arise commonly in wave an heat theory. Many ientities for them look similar to ientities for the orinary trigonometric functions cos an sin, but sometimes with a change of sign. Important ientity: cosh x sinh x This looks like the well known trigonometric ientity cos x + sin x, but note that there is a change of sign. The above ientity can easily be erive from the basic efinitions () as follows: ( e cosh x sinh x + e x ) ( e x e x ) x Another ientity: ex + + e x 4 ( ) (ex + e x ) 4 sinh(x + y) sinh x cosh y + cosh x sinh y () This, too, can be erive from the basic efinitions (): ( e x e x ) ( e y + e y ) ( e x + e x ) ( e y e y ) RHS of () + ex+y + e x y e x+y e (x+y) 4 ex+y e (x+y) sinh(x + y) 0. The hyperbolic tangent function: tanh The hyperbolic function tanh x is efine by tanh x sinh x cosh x + ex+y e x y + e x+y e (x+y) 4
tanh x is an o function. As in orinary trigonometry, if we know the sinh or cosh of a number we can work out the other hyperbolic functions of that number, as the following example emonstrates. 0.3 Example If sinh x 8 calculate cosh x an tanh x. 5 Solution. We know that cosh x sinh x. Lets use this ientity to fin cosh x. We have ( ) 8 cosh x + sinh x + 89 5 5 From the basic efinition of cosh in () we see that the cosh of anything is always positive. Hence cosh x 7. 5 Finally, tanh x sinh x 8/5 8. cosh x 7/5 7 0.4 Example Solve for x the equation 5 cosh x + 3 sinh x 4. Solution. From the basic efinitions of sinh an cosh, the equation can be written as ( e x + e x ) ( e x e x ) 5 + 3 4 which simplifies to 4e x + e x 4. Multiplying by e x gives 4e x + 4e x, which can be written in the form of a quaratic equation 4(e x ) 4e x + 0 From the formula for solving a quaratic equation, we fin that e x 8 i.e. e x. Hence x ln ln ln. 0.5 An ientity for tanh tanh(x + y) ( 4 ± 0 ) tanh x + tanh y + tanh x tanh y This looks a bit like the corresponing ientity for the orinary trigonometric function tan, namely: tan x + tan y tan(x + y) tan x tan y but note there is a ifference in sign in the enominator. Let us erive formula (3). We alreay have sinh(x + y) sinh x cosh y + cosh x sinh y (3) 3
an we can also show that Hence cosh(x + y) cosh x cosh y + sinh x sinh y tanh(x + y) sinh(x + y) cosh(x + y) sinh x + sinh y cosh x cosh y sinh y + sinh x cosh x cosh y tanh x + tanh y + tanh x tanh y sinh x cosh y + cosh x sinh y cosh x cosh y + sinh x sinh y As commente on previously, ientities for hyperbolic functions often look like those for the orinary trigonometric functions sin, cos, tan, but there is often a change of sign. There is a general rule for eriving an ientity for hyperbolic functions from the corresponing ientity for orinary trigonometric functions. It is calle Osborn s rule. 0.6 Osborn s rule To get a formula for hyperbolic functions from the corresponing ientity for orinary trigonometric functions, replace every orinary trigonometric function by the corresponing hyperbolic function, an change the sign of every prouct or implie prouct of sine terms. Implie prouct means things like sin x which can be written as sin x sin x an is therefore a prouct of sines. Example. We know that cos(x + y) cos x cos y sin x sin y. Replace cos by cosh an sin by sinh. Also sin x sin y is a prouct of sines so change the sign from to +. In this way we erive the ientity cosh(x + y) cosh x cosh y + sinh x sinh y Example. We know that cos x + sin x. Now sin x sin x sin x so is a prouct of sine terms. We write the above ientity as cos x cos x + sin x sin x Now change cos to cosh, sin to sinh, an change the sign of the sin x sin x term which is a prouct of sines. We erive the ientity Example. We know that cosh x sinh x tan(x + y) tan x + tan y tan x tan y 4
which can be written as sin x sin(x + y) cos(x + y) + sin y cos x cos y sin x sin y cos x cos y In the above expression there appears the prouct sin x sin y which is a prouct of sines so its sign shoul be change. Also, replace sin by sinh an cos by cosh to erive the following: tanh x + tanh y tanh(x + y) + tanh x tanh y 0.7 Derivatives of hyperbolic functions It can be straightforwarly shown from the basic efinitions () that (sinh x) x cosh x (cosh x) x sinh x x (tanh x) sech x The thir of these can be erive from the quotient rule for erivatives: ( ) sinh x (tanh x) x x cosh x cosh x sinh x cosh x cosh x sech x 0.8 Other hyperbolic functions an their erivatives The function coth x is efine by coth x cosh x sinh x an is analogous to the orinary trigonometric function cot x, which equals cos x/ sin x. The erivative of coth x can be foun using the quotient rule as follows: ( ) cosh x (coth x) x x sinh x sinh x cosh x sinh x sinh x cosech x 5
where The function sech x is efine by cosech x sinh x an its erivative is given by sech x cosh x The function cosech x is efine by (sech x) sech x tanh x x an its erivative is given by cosech x sinh x 0.9 Example (cosech x) cosech x coth x x Fin t (tanh + t ). Solution. To o this we shall have to use the chain rule. Let u + t. Then t (tanh + t ) (tanh u) ( t ) u (tanh u) u t (sech u) ( + t ) / (t) t sech + t + t 0.0 Integrals of hyperbolic functions sinh x x cosh x + c cosh x x sinh x + c sech x x tanh x + c 6
0. Example Fin coth 5x x. Solution. coth 5x x cosh 5x x sinh 5x let u sinh 5x so u 5 cosh 5x x /5 u u u u 5 0. Example Fin sinh x x. 5 ln u + c ln(sinh 5x) + c 5 Solution. Nee to re-write sinh x as something that can be more easily integrate. Start with the ientity cosh(a + B) cosh A cosh B + sinh A sinh B Put B A in this to get cosh A cosh A + sinh A. Recall also that cosh A sinh A. Putting these things together gives cosh A + sinh A so that Hence sinh A (cosh A ) sinh x x (cosh x ) x ( ) sinh x x + c 0.3 Inverse hyperbolic functions If y sinh x then we say x sinh y, the inverse hyperbolic sin of y. NB: sinh y is just notation for the inverse sinh of y. It oes not mean the same as (sinh y). The function y cosh x is not one to one. This is because each y value has two corresponing x values. However, if we restrict to x 0 then the function cosh x oes have an inverse, written cosh x. 7
On moern calculators inverse hyperbolic functions are usually accesse using a shift an a hyp button. A typical sequence for sinh might be something like shift hyp sin. 0.4 Derivatives of inverse hyperbolic functions Let y cosh x an suppose we want to fin y/x. Since y cosh x, it follows that cosh y x. Implicit ifferentiation of this equation gives sinh y y x so that y x sinh y. But recall that cosh y sinh y. Using this fact gives y x sinh y cosh y x We have shown that Similarly, an x (cosh x) x (sinh x) x x + x (tanh x) x 0.5 Example Evaluate 0 x. 3 + 4x Solution. The appearance of the integran suggests that we shoul aim to use the result x + x sinh x + c but we shall first nee to make a substitution. Note that 3 + 4x 3 ( + 4 3 x) ; this suggests we shoul use the substitution x 3y, so that x 3 y. Making this substitution, we get / 3 x ( ) 3 y 0 3 + 4x 0 3 + 4 ( ) 3 4 y / 3 y 3 0 3 + 3y 8
0.6 Example Fin x x. / 3 y 3 0 3 + y [ sinh y ] / 3 sinh 3 sinh 0 0.9866 Solution. Substitute x y. Then x y. We have x y x y 0 y y y y cosh y + c cosh x + c 0.7 Example Fin a + x x. Solution. Recall cosh θ sinh θ, so that cosh θ + sinh θ. This suggests we shoul substitute x a sinh θ in the integral. Carrying out the substitution gives a + x x a + a sinh θ a cosh θ θ a + sinh θ a cosh θ θ a cosh θ θ We may apply Osborn s rule to the orinary trigonometric ientity cos θ + cos θ, to euce that cosh θ + cosh θ Therefore a + x x a ( + cosh θ) θ ( ) sinh θ a θ + + c a (θ + sinh θ cosh θ) + c 9
0.8 Example Fin Solution. x x 6x + 5. x x 6x + 5 a sinh x a + x + x + c a a x (x 3) 4 t 4t 4 cosh t + c ( ) x 3 cosh + c t t Substitute x 3 t 0