1. Hyperbolic functions

2. FM Hyperbolic functions:
KUS objectives
BAT use the inverse hyperbolic functions
Starter: Expand and simplify
=ex+y
ex x ey
(e-x – 1)2
=e−2x – 2e−x + 1
=¼(e2x – 2 + e−2x)
[½(ex – e-x)]2

3. If you take a rope/chain, fix the two ends, and let it hang under the force of gravity, it will naturally form a hyperbolic cosine curve.
Hyperbolic functions
Most curves that look parabolic are actually Catenaries, which is based in the hyperbolic cosine function. A good example of a Catenary would be the Gateway Arch in Saint Louis, Missouri.

4. tanh 𝑥 ≡ sinh 𝑥 cosh 𝑥 ≡ 𝑒 2𝑥 −1 𝑒 2𝑥 +1
Notes Hyperbolic functions have several properties in common with trigonometric functions, but they are defined in terms of exponential functions
sinh 𝑥 ≡ 𝑒 𝑥 − 𝑒 −𝑥 2
cosh 𝑥 ≡ 𝑒 𝑥 + 𝑒 −𝑥 2
tanh 𝑥 ≡ sinh 𝑥 cosh 𝑥 ≡ 𝑒 2𝑥 −1 𝑒 2𝑥 +1
‘shine x’
‘coshine x’
There are corresponding reciprocal functions
cosech 𝑥 ≡ 2 𝑒 𝑥 − 𝑒 −𝑥
sech 𝑥 ≡ 2 𝑒 𝑥 + 𝑒 −𝑥
coth 𝑥 ≡ 1 tanh 𝑥 ≡ 𝑒 2𝑥 +1 𝑒 2𝑥 −1
‘cosheck x’
‘sheck x’
These definitions can simply be stated but need to be memorised

5. WB A1 Find to 2dp the values of 𝑎) sinh 5 b) cosh ( ln 2 ) 𝑐) tanh 𝑥 2
tanh 𝑥 ≡ sinh 𝑥 cosh 𝑥 ≡ 𝑒 2𝑥 −1 𝑒 2𝑥 +1
a) sinh 5 ≡ 𝑒 5 − 𝑒 −5 2 = 74.20
b) cosh ( ln 2 ) ≡ 𝑒 ln 2 + 𝑒 −𝑙𝑛2 2 = = 5 4
c) tanh 𝑥 2 ≡ 𝑒 2 𝑥 2 −1 𝑒 2 𝑥 = 𝑒 𝑥 −1 𝑒 𝑥 +1

6. WB A2 Find the values of x for which 𝑎) sinh 𝑥 =5 𝑏) tanh 𝑥 = 15 17
cosh 𝑥 ≡ 𝑒 𝑥 + 𝑒 −𝑥 2
tanh 𝑥 ≡ sinh 𝑥 cosh 𝑥 ≡ 𝑒 2𝑥 −1 𝑒 2𝑥 +1
a) sinh 𝑥 ≡ 𝑒 𝑥 − 𝑒 −𝑥 2 =5
multiply through by 𝑒 𝑥 to get a quadratic
Rearrange to 𝑒 𝑥 − 𝑒 −𝑥 −10=0
Use quadratic formula to solve
𝑒 2𝑥 −10 𝑒 𝑥 −1=0
𝑒 𝑥 = −10± (−10) 2 −4(−1) 2 = =2.31
b) tanh 𝑥 ≡ 𝑒 2𝑥 −1 𝑒 2𝑥 +1 = 15 17
17( 𝑒 2𝑥 −1) =15( 𝑒 2𝑥 +1)
𝑒 2𝑥 =32
2𝑥= ln 32 =4 ln so 𝑥=2 ln 2

7. Graphs

8. WB3 sketch the graphs of sinh 𝑥 , cosh 𝑥 𝑎𝑛𝑑 tanh 𝑥

9. WB3a (cont) sketch the graphs of sinh 𝑥 , cosh 𝑥 𝑎𝑛𝑑 tanh 𝑥
tanh 𝑥 ≡ sinh 𝑥 cosh 𝑥 ≡ 𝑒 2𝑥 −1 𝑒 2𝑥 +1
the graph of sinh x is the ‘average’ of the graphs of 𝑦=𝑒 𝑥 and y=− 𝑒 −𝑥
the graph of sinh x is an odd function
sinh (−𝑥) =− sinh 𝑥
Note that:
As 𝑥→∞ ,− 𝑒 −𝑥 →0 so sinh x≈ 1 2 𝑒 𝑥
And
As 𝑥→−∞ , 𝑒 𝑥 →0 so sinh x≈ − 1 2 𝑒 −𝑥
Domain 𝑥∈𝑅
Range 𝑦 ∈𝑅

10. WB3b (cont) sketch the graphs of sinh 𝑥 , cosh 𝑥 𝑎𝑛𝑑 tanh 𝑥
tanh 𝑥 ≡ sinh 𝑥 cosh 𝑥 ≡ 𝑒 2𝑥 −1 𝑒 2𝑥 +1
the graph of cosh x is the ‘average’ of the graphs of 𝑦=𝑒 𝑥 and y= 𝑒 −𝑥
the graph of cosh x is an even function
cosh (−𝑥) = cosh 𝑥
Note that:
As 𝑥→∞ , 𝑒 −𝑥 →0 so cosh x≈ 1 2 𝑒 𝑥
And
As 𝑥→−∞ , 𝑒 𝑥 →0 so cosh x≈ 1 2 𝑒 −𝑥
Domain 𝑥∈𝑅
Range 𝑦 ≥1

11. WB3c (cont) sketch the graphs of sinh 𝑥 , cosh 𝑥 𝑎𝑛𝑑 tanh 𝑥
tanh 𝑥 ≡ sinh 𝑥 cosh 𝑥 ≡ 𝑒 2𝑥 −1 𝑒 2𝑥 +1
Consider the graphs of cosh x and sinh x to understand the graphs of tanh x
tanh 𝑥 ≡ sinh 𝑥 cosh 𝑥
the graph of tanh x is an odd function
tanh (−𝑥) = 𝑡𝑎𝑛h 𝑥
Note that:
As 𝑥→∞ , tanh 𝑥 →1
And
As 𝑥→−∞ , tanh 𝑥 →−1
Domain 𝑥∈𝑅
Range −1<𝑦 <1

12. Any comments about these answers?
WB A4 Evaluate the following, leaving answers to 4 s f a) sinh 7 b) cosh(-5) c) tanh 0.2 Solve the hyperbolic equations, leaving answers correct to 3sf d) sinh x = -2 e) cosh x = 3 f) tanh x = 0.8 Try finding the exact solutions using the definitions for sinh, cosh and tanh. Compare your answers to those in question 2. Any further comments now?
548.3
74.21
0.1974
coshx is a many-one function and therefore two solutions are required. Calculators only give the principal value.
Any comments about these answers?
x = -1.44
x = 1.76
x = 1.10
Y = sinhx and tanhx are one-one functions and therefore there is just the one answer
Y = coshx is a many-one function and therefore two solutions were/are required. Calculators give the principal value.
𝑦=𝑠𝑖𝑛ℎ𝑥
𝑦=𝑡𝑎𝑛ℎ𝑥
𝑦=𝑐𝑜𝑠ℎ𝑥

13. NOW DO EX 6A WB A5 sketch the graphs of cosech 𝑥 , 𝑠𝑒𝑐h 𝑥 𝑎𝑛𝑑 coth 𝑥
y=cosech 𝑥
y=sinh 𝑥
y=coth 𝑥
y=tanh 𝑥
y=sech 𝑥
y=cosh 𝑥
NOW DO EX 6A

14. One thing to improve is –
KUS objectives
BAT use the inverse hyperbolic functions
self-assess
One thing learned is –
One thing to improve is –

15. END