I Relate the Hyperbolic Functions to the Exponential Function I Sketch Graphs

In th is ch apter you will learn how to:

I relate the hyperbolic functions to the exponential function I sketch graphs of the hyperbolic functions I prove and use identities involving hyperbolic functions I use the definitions of the inverse hyperbolic functions and use the logarithmic forms. Cambridge International AS & A Level Further Mathematics: Further Pure Mathematics 2 ; '~

PREREQUISITE KNOWLEDGE

Where it comes from What you should be able to do Check your skills AS & A Level Mathematics Use the trigonometric identities. 1 Prove: Pure Mathematics 2 & 3, cos A sin A cos(A + B) Chapter 3 ------sinB cosB sinBcosB

AS & A Level Mathematics Manipulate equations involving 2 Solve: e2x - e - 6 = 0 Pure Mathematics 2 & 3, the natural logarithm and Chapter 2 exponential functions.

What are hyperbolic functions? In the Pure Mathematics 2 & 3 Coursebook, Chapter 4, you learned uses for trigonometric functions other than solving problems based on a triangle. In fact, trigonometric functions are also called circular functions, as they all have geometric meaning and derivations from the equation of the unit circle. There are similar relationships between the structure of circles, hyperbolas, parabolas and ellipses. These shapes are all called conic sections, because they are shapes produced when we slice a cone. Horizontal slices produce circles; oblique slices produce ellipses; slices parallel to the slope of the cone produce parabolas; steeper slices than this produce hyperbolas.

circle ellipse parabola hyperbola

Hyperbolic functions are a class of functions that are derived from the unit hyperbola, 2 2 x - y = 1. They are a significant tool in an engineer's toolkit. They also allow us to find highly elegant solutions to differential equations.

19.1 Exponential forms of hyperbolic functions In the introduction, we discussed the relationships between hyperbolic functions and trigonometric functions. Hyperbolics link to apparently vastly different function types, but as you read this chapter and do subsequent work on complex numbers, you will start to see the connections.

We shall start by relating hyperbolic functions to exponentials. It is not clear from the ideas and development of hyperbolics whether the exponential definitions came first or the relationships with trigonometry came first. We shall use the exponential forms to develop the idea. Let us begin with the hyperbolic sine function: y = sinhx I. . Chapter 19: Hyperbolic functions

The first challenge is the pronunciation of the hyperbolic functions. The hyperbolic sine is ' ] pronounced 'shine' and its exponential definition is: . ex - e-x smhx = 2

Based on this, we can define the domain, find the range and draw the function. It is helpful to consider this graph as the average of y = e and y = -e-x.

------5 -4 - 3 - 2

-4

From this, we can see that the function is a one- one mapping. The domain is x E IR.. The range is f(x) E IR.. It is also an odd function.

y = coshx The hyperbolic cosine is pronounced 'cosh' and its exponential definition is: ex+ e-x coshx=-- - 2

We consider this graph as the average of y =ex and y = e-x .

------...... ------.. -- - 5 -4 - 3 - 2 - 1 0 1 2 3 4 x

::-rom this, we can see that the function is not a one- one mapping. However, it can be .valuated with both vositive and negative values for x.

the domain is x E IR.. The range is f( x) ~ 1. It is an even function.

'Or the function to be one- one, we need to restrict the domain to x ~ 0. This will be very mportant later. Cambridge International AS & A Level Further Mathematics: Further Pure Mathematics 2 J

0 DID YOU KNOW? The hyperbolic cosine is a very special curve and has many occurrences in nature. It is called a catenary, and is the shape of a hanging chain.

EXPLORE 19.1

Based on the exponential definitions of sinhx and coshx, find:

~(sinhx) and ~(cosh x) dx dx

The exponential forms of sinh x and cosh x are shown in Key point 19.1.

KEY POINT 19.1

The exponential forms of sinh x and cosh x: . ex - e- x smh x = , x E IR. , f(x) E IR. 2 ex + e-x I cosh x = , x E IR., f(x) ~ I 2 From this, we can deduce that ex = cosh x + sinh x.

We can now derive further hyperbolic functions:

Hyperbolic function Pronunciation Exponential form tanhx 'tanch' (preferred) ex - e-x tanhx= - - - 'than' (a long thhhhhh as in 'thank you') ex + e-x 'tank'

Y ,~ Domain: x E lRI. 3 2

1 / Range: -1 < f(x) < 1 ..... ~ -4 - 3 - 2 0 1 2 3 4 x Y.- 1 - 2 Odd function - 3 I. Chapter 19: Hyperbolic functions

erbolic function Pronunciation Exponential form 2 'setch' sechx=-- ex + e-x y Domain: x E !R 2

Range: 0 < f(x) ~ 1

-4 - 3 - 2 - 1 2 3 4 x - 1 Even function - 2

DID YO~ KNOW?

The graph of y = sech2 x is a very important curve. It is used to model tsunamis and to model wavelets in wave- particle duality. We call this type of wave a soliton. It is a very special type of wave that does not lose energy as it travels. Its velocity is proportional to the amplitude of the wave. This is why tsunamis have such a destructive force. You might study more about solitons at university level.

yperbolic function Pronunciation Exponential form 2 cosechx 'cosetch' cosech x = --- ex - e-x Domain: x-=I= 0

Range: f(x) -::fa 0

4 x

Odd function

coth x

Domain: x -=I= 0 To help us remember the reciprocal functions of sech x, cosech x and coth x, we can apply the third letter rule Range: f(x) > 1, f(x) < -1 as with trigonometric reciprocal functions: -4 - 3 - 2 0 1 2 3 4 x 1 se chx= - - coshx Odd function 1 cosechx = -- sinhx 1 cothx= -- tanhx Cambridge International AS & A Level Further Mathematics: Further Pure Mathematics 2 .~ t'

WORKED EXAMPLE 19.1

Sketch the graph of y = l3cosh(x + 1) - 41. Answer

Given that f(x) = cosh x : Consider each transformation in turn.

I I y = f(x + 1) I I I I I I I I I I I I I I -"

2 3 4 x

I I I I y=3f(x +l) I I I I I I I I I I I I I I I I \ I \ I \ I \ I \ I \ I \ 2 I ''--!''

I -4 - 3 - 2 - 1

y=3f(x +l) -4

This is the graph of 13 cosh(x + 1) - 41. y = l3f(x + 1) - 41

-4 - 3 2 3 4 x i Chapter 19: Hyperbolic functions

These functions can all be evaluated using a calculator, in the same way as for trigonometric functions. However, generally you will be required to write in exact form, unless told otherwise. We can use the previous definitions to solve equations involving hyperbolic forms.

~- WORKED EXAMPLE 19.2 & Leaving your answer in exact form, solve 4coshx - 3 sinh x = 4. Answer

4cosh x - 3 sinh x = 4 First, use the exponential definitions: . ex - e-x 4 (e' ~c x ) - 3(e' ~ e-x) = 4 smh x= --- 2 ex + e-x coshx= --- 2

4 e2-' + 4 - 3e 2x + 3 = 8 ex Now multiply up by 2 ex.

e2x - 8 ex + 7 = 0 This is now a quadratic in terms of e and so can be solved. ( e' - 7)(e -' - 1) = 0

Leading to two solutions: x = ln 7, x = In 1 = 0 We are usually asked to give our solutions in exact form.

It is possible to approach this problem using a similar technique as that for trigonometric equations using hyperbolic forms. However, this rarely gives such elegant solutions. We need to understand both inverse hyperbolic functions and their logarithmic forms.

Solve 3 tanh2 x - 4 tanh x = 4. Answer

3tanh2x - 4tanhx - 4 = 0 First, recognise that this is a quadratic equation in terms of tanh x.

To make it simpler, replace tanh x with T

3T2 -6T+ 2T-4=0 Solve the equation. 3T(T - 2) + 2(T - 2) = 0 (T - 2)(3T + 2) = 0 T = 2 2 Or T = - - 3 We are left then to solve: Remember: since the range of tanh x ) tanh x = -=- cannot exceed 1, T = 2 will not lead to any 3 solutions. x = tanb- 1 (-D ~ Cambridge International AS & A Level Further Mathematics: Further Pure Mathematics 2 · j

Write all solutions in exact form unless otherwise stated.

1 Solve 19sinhx + 16coshx = 8.

2 Solve 29coshx = 11 sinhx + 27.

3 Solve 17 sinhx + 16 coshx = 8.

4 Solve 7 = 17 tanhx + 28 sechx.

5 Solve cosechx - 2cothx = 2.

6 Sketch the graph of y = 13 coshx - 51.

7 Sketch the graph of y = 4 - cosech(x - 2).

8 Sketch the graph of y = 12 + _!_coth(x + 1)1. 2

~ 9 Solve 4coshx + sinhx = 8. Write your answers in the form b In a, where a and bare integers.

~ 10 Solve lOcoshx + 2sinhx = 11, giving your answers in the form ln a, where a is a rational number.

~ 11 Given that sech-1 x = In (1 +?),solve 4sechx - 3tanh2x =I. I. 19.2 Hyperbolic identities If we consider that hyperbolic functions are similar to trigonometric functions, it is reasonable to consider that hyperbolic identities exist in the same way. We can use the exponential forms to prove that they are true. Q __ I When forming a proof, WORKED EXAMPLE 19.4 it is very important to either start at the 2 2 left-hand side and Prove that cosh x - sinh x = 1. show it is the same as Answer the right-hand side, or 2 2 start at the right-hand cosh x - sinh x side to show it is the same as the left-hand F irst, convert to exponential fo rm and side. Starting from consi der the left-hand side. both ends until you get a common expression 1 will not make a formal = _!_ ( ( e2x + 2 + e - 2x) _ ( e 2x _ 2 + e- 2')) Expand and take out a factor of - . 4 4 proof. You can use this to help you see how to = _!_(e2.\'. + 2 + e-2x _ e 2x + 2 _ e - 2x) 4 formulate your proof, but the final written _!_(4) 1 This is the right-hand side. answer must be in the =4 = correct order. Make Therefore, cosh2 x - sinh2 x = 1. Therefore, the left-hand side becomes the sure that you state right-hand si de. every algebraic step: tell the whole story! I Chapter 19: Hyperbolic functions

~' WORKED EXAMPLE 19.5 tm Prove that sinh(A + B) = sinh A cosh B + sinh B cosh A . Answer si nh A cosh B + sinh B cosh A

Consider the right-hand side and convert it to exponential fo rm.

1 Expand and take out a factor of -. 4

e(A+ B) - e - (A + B) = = sinh(A + B) T his proof works from right-hand side to 2 left-hand si de. as required.

You should have noticed that these two proofs are similar to trigonometric identities. It is possible to use Osborne's rule, as shown in Key point 19.2, to move from a trigonometric identity to a hyperbolic identity. Nevertheless, this does not constitute a proof!

Osborne's rule To move fro m a trigonometric iden tity to a hyperbolic identity: change a cos to a cosh and change the sign of a product of sinhs.

We can use Osborne's rule when we are required to use a hyperbolic identity without needing its proof.

Using Osborne's rule, state the angle addition formula for tanh(A + B). Answer

Start by considering the trigonometric identity: We consider replacing tan with tanh.

tan(A + B) = tanA + tanB 1 - tanA tanB

tanh(A + B) = tanh A + tanh B sinh is the numerator of tan h. Applying 1 tanh Atanh + B Osborne's rule, we notice that as required. tanhAtanhB will require a sign change since it includes a product of sinhs. cosech2x + 1 1 Prove that cos h2x =----- G coth2 x - 1 G 2 Prove that cosh3x =4cosh 3 x - 3 coshx. 1 3 Prove that coth2x - cothx =-(tanhx - cothx). G 2 G 4 Prove the following hyperbolic identities. a cosh(A - B) =coshA coshB - sinhA sinhB b sinh 3x = 3 sinhx + 4 sinh3 x

c sinhA - sinhB =2cosh( A; B)sinh( A; B)

5 Given that tanh (~) = t, prove that sinh x = __}!____ 2 1 - t2

19.3 Inverse hyperbolic functions As with trigonometric functions, the hyperbolic inverse functions must exist and be well defined. As with any function, the inverse function will be a reflection in the line y = x for the one-one aspect of the function (otherwise the inverse will not be well defined). I Take care to use the correct notation. You may have noticed by now that the notation for inverse trigonometric functions is a little ambiguous and it can be easily confused with reciprocal functions. There is another commonly used notation for them: sin-1 x = arcsin x cos-1 x = arccosx tan-1 x = arctan x 0 DID YOU KNOW? . The origin of the prefix arc- comes from the unit circle, and the fact that the arc length is equal to the angle (l =re, where r = 1). Finding the angle e is then equivalent to finding the arc length. So finding the angle, which requires inverting the sine function, is equivalent to finding the arcsine.

There is a similar idea with hyperbolics, although beyond the scope of this course. Finding the hyperbolic angle is equivalent to finding twice 2 2 the area of the hyperbolic sector of x - y = 1, the unit hyperbola. I Chapter 19: Hyperbolic functions

-!KEY POINT 19.3 T-he inverse hyperbolic functions are: Domain and range of the inverse function x E IR f(x) E IR.

1 cosh x cosh- x x~ I f(x) ~ 0

tanhx tanb- 1 x - 1

Solve the equation sinh2 x- 5sinhx + 4 = 0. Give your answers to 3 significant figures. Answer sinh2x - 5sinh x + 4 = 0 First, notice that this is a quadratic equation in terms of (sinh x - 4) (sinh x - 1) = 0 sinhx, and it will factorise.

When sinh x = 4, x = sinh- 14: Make sure solutions are written to the required level of x == 2.09 accuracy. When sinh x = 1, x = sinh- 1 1: x == 0.88 1

It is Possible to write the answer exactly, using natural logarithms. 19.4 Logarithmic form for inverse hyperbolic functions If hyperbolic functions can be described in terms of exponential functions, then it is reasonable to assume that inverse hyperbolic functions can be described using the natural logarithm function. T his is indeed the case and is shown in Key point 19.4.

I WORKED EXAMPLE 19.8

Find the logarithmic form of sinh-1 x.

Answer

y = sinh- 1 x sinhy = x Rewrite x in terms of sinh y. eY - e-Y - - - = x Express in exponential form. 2

e2Y - 1 = 2eYx Rearrange to ma ke y the subject. e2Y - 2eYx - 1 = 0 2 ( eY - x )2 - x - 1 = 0

2 2 ( eY - x ) = x + 1 To do that here, complete the square. eY= x ± R+l

y = ln(x + Vx 2 + 1) Take logs to m ake y the subject.

Therefore, sinh- 1 x = ln(x + R+l). Since we can only take logs of positive numbers, take only the positive root here. Consider why we do this rather than take the modulus.

KEY POINT 19.4

The logarithmic form s of the inverse hy perbolic functions are: Why do we consider only the positive root sinh- 1 x 2 ln(x + V x + 1) here?

1 2 cosh- x ln(x + V x - 1)

tanh- 1 x l1n2 ( ~1 - x ) 1 coth- 1 x l 1n(x + ) 2 x - l

sech-1 x ln(' + v; -x' )

cosech- 1 x ln(' + v; + x' )

You need to be able to derive the logarithmic form for inverse hyperbolics.

You are now equipped to give your answers in exact fo rm. In the next example, we will solve the equation we had in Worked example 19.7 but this time we will give our answers in exact form. I Chapter 19: Hyperbolic functions

"'ftl_ WORKED EXAMPLE 19.9 ~·

Solve the equation sinh2 x - 5sinh x + 4 = 0. Give your answers in exact logarithmic fo rm. Answer

sinh2 x - 5sinhx + 4 = 0 Notice that this is a quadratic equation in terms of (si nh x - 4) ( sinh x - 1) = 0 sinh x, and it will factorise.

When sinh x = 4, x = sinh- 1 4: sinh-14 = ln(4 + v' 42 + 1) x = ln(4 + m) When sinh x = 1, x = sinh- 1 1: sinh- 1 1 = ln(l + ~) x = ln(l + v'2) Note the answer has been requested in exact logarithmic form .

~-' --- -,. WORKED EXAMPLE 19.10 ~ ~!.... .' .• J,.

Prove that sinh- 1 A + sinh- 1 B = sinh- 1(AV1 + B 2 + B~) .

Answer

2 2 2 2 2 2 1 + (A V l + B + BVl + A ) Consider 1 + (A v' 1 + B + BV1 + A ) .

= 1+ 2A 2B 2 + 2A Bv'A2+J VB 2 + 1 + A2 + B 2

2 2 2 2 2 2 2 2 2 2 2 2 = A B + 2ABV A + 1V B + 1 + (A + l)(B + 1) (1 + A )(1 + B ) =1 + A + B + A B

= (AB + v' A2 + 1~ ) 2 CalJ this result (1).

1 2 2 sinh- (AV1 + B + BV l + A ) =: ln (A ~ +B~ + Vl+ (A~ +B~) 2 ) =l n(A V l + B 2 + BV l + A 2 + AB + v' A 2 + 1 v' B 2 +I) Use result (1). :::: ln((A + V 1 + A 2)(B + ~) ) T his is of the form AC + BD + AB + CD . It will factorise to (A + D)(B + C).

2 2 =: ln(A + V l + A ) + ln(B + V l + B ) Rewrite, using the laws of logarith ms.

== sinh- 1 A + sinh- 1 B as required .

There are some very interesting relationships between the inverse hyperbolic functions. We do not need to know these but they will help with some of the hyperbolic and trigonometric integration later.. For example:

sinh(cosh- 1 x) = Vx2=J 2 1 x - In x = tanh- - --I) ( 1 x- + I Cambridge International AS & A Level Further Mathematics: Further Pure Mathematics 2 l

1£HMH!i'* 1 Find the exact value of:

1 - 1 2 1 a sinh- 3 b tanh - c cosh- ~ 3 4 2 Solve, giving your answers as logarithms. a 3 cosh2 x = -10 sinhx b tanh 2 x + 5 sech x - 5 = 0 c sinh2x = coshx G 3 Prove that cosh-1 x = ln(x + ~).

4 Prove that tanh- 1 x =-I ln ('+x)-- . G 2 1-x

G 5 Prove that sech-1 x = ln ('+~)x .

1 1 1 6 Prove that tanh- A+ tanh- B =tanh- ( A+B) . G 1 + AB G 7 a Prove that 4cosh3x - 3coshx =cosh3x. I b Hence, solve cosh 3x = 5 cosh x. WORKED EXAM-STYLE QUESTION

Given that sinh y = x, show that: a y =ln(x+ ~) By differentiating this result, show that: b dy dx ~ Answer

a sinhy = x Use the exponential definition. eY - e- .v - --= x 2 e2Y - 1 = 2e Yx 2 e Y - 2e Yx - 1 = 0

2 2 ( e Y - x ) - x - 1 = 0 Complete the square. (e .J' - x )2 = x 2 + 1

eJ' =x±R+!

y = In(x + Vx 2 + 1) Rearrange and take the positive root. I Chapter 19: Hyperbolic functions .

b Let y = ln u. dy Therefore, - du u u = x +(I + x 2) ~

Therefore, du= 1 + (!) 2x(l + x 2t t Simplify. dx 2

= 1 + --x__ ~ ~ + x ~ dy dy du - = - X Use the chain rule. dx du dx dy , v1 + x 2 + x I ----- x - ---- Use u = x +(I + x2)2. dx x +(l + x 2) ~ ~ dy ---1 as reqmre. d . dx v1 + x 2

For hyperbolic functions: . ex - e-x • smhx = -- 2 e-' + e-x • coshx = -- 2

e-' - e-x e1x - 1 • tanhx=-- = --· 2 e + e-x e X + 1 For inverse functions: • arcsinhx = sinh- 1 x = Jn(x + Vx 2 + 1) • arccosh x = cosh- 1 x = ln(x + ~)

• arctanhx = tanh- 1 x = -1ln (l-- + x) 2 1 - x 0 1 Solve 5coshx- cosh2x = 3. Leave your answer in exact logarithmic form. 0 2 Starting from the definitions of sinhx and coshx in terms of exponentials, prove that: a 1+2sinh2x =cosh2x b 2cosh2x + sinhx = 5 0 3 a Show, by using exponential form, that the curve with equation y = cosh2x + sinhx has exactly one stationary point. b Determine, in exact logarithmic form, the x-coordinate of the turning point.