1. FP3 Chapter 1 Hyperbolic Functions
Dr J Frost
Last modified: 15th November 2016

2. Recap of Conic Sections
The axis of the parabola is parallel to the side of the cone.
All the ones you’ll see can be obtained by taking ‘slices’ of a cone (known as a conic section).
C2:
Circles
FP1:
Rectangular Hyperbolas
FP3
Ellipses
FP1:
Parabolas
FP3:
Hyperbolas

3. Comparing different conics
No need to make notes on this yet. We’ll cover it in Chapter 2.
We saw circles and ‘rectangular hyperbolas’ back in FP1.
Parabolas
Circles
Hyperbolas
𝑥=−𝑎 𝑦
𝑃(𝑥,𝑦) 𝑎 𝑋 𝑆 𝑥
Picture: Wikipedia 1
𝑥,𝑦 𝜃 1
The ‘simplest’ circle is a unit circle centred at the origin.
Cartesian equation:
𝒙 𝟐 + 𝒚 𝟐 =𝟏
Parametric eqns (in terms of 𝜃):
𝒙=𝒄𝒐𝒔 𝜽 𝒚=𝒔𝒊𝒏 𝜽
The equivalent hyperbola (which crosses 𝒙-axis at (𝟏,𝟎) and −𝟏,𝟎 )
Cartesian equation:
𝒙 𝟐 − 𝒚 𝟐 =𝟏
Parametric equations:
𝒙=𝒄𝒐𝒔𝒉 𝜽 𝒚=𝒔𝒊𝒏𝒉 𝜽
2 focal points and 2 directrices.
Distance to focus and distance to (nearest) directrix can differ by constant factor (known as eccentricity)
If 𝒂=𝟏:
Cartesian equation:
𝒚 𝟐 =𝟒𝒙
Has 1 focus and
1 directrix.
Important property?
Distance from focus and directrix equal. ?
similar ? ? ?
similar ? ?

4. What’s the point of hyperbolas?
Hyperbolic functions often result from differential equations (e.g. in mechanics), and we’ll see later in FP3 how we can use these functions in calculus.
For example, we can consider forces acting on each point on a hanging piece of string.
Solving the relevant differential equations, we end up with:
𝒚= 𝐜𝐨𝐬𝐡 𝒙 ?
OMG modelling!

5. Equations for hyperbolic functions
Hyperbolic sine:
sinh 𝑥 = 𝑒 𝑥 − 𝑒 −𝑥 𝑥∈ℝ
Hyperbolic cosine:
cosh 𝑥 = 𝑒 𝑥 + 𝑒 −𝑥 𝑥∈ℝ
Hyperbolic tangent:
tanh = sinh 𝑥 cosh 𝑥 = 𝑒 2𝑥 −1 𝑒 2𝑥 𝑥∈ℝ
Hyperbolic secant:
sech 𝑥 = 1 cosh 𝑥 = 2 𝑒 𝑥 + 𝑒 −𝑥 𝑥∈ℝ
Hyperbolic cosecant:
cosech 𝑥 = 1 sinh 𝑥 = 2 𝑒 𝑥 − 𝑒 −𝑥 𝑥∈ℝ, 𝑥≠0
Hyperbolic cotangent:
coth 𝑥 = 1 tanh 𝑥 = 𝑒 2𝑥 +1 𝑒 2𝑥 − 𝑥∈ℝ, 𝑥≠0 ?
Say as “shine” ?
Say as “cosh”
Say as “tanch” ? ?
Say as “setch” ? ? ? ?
Say as “cosetch” ? ?
Say as “coth”

6. Equations for hyperbolic functions
Calculate (using both your 𝑠𝑖𝑛ℎ button and using the formula)
𝐬𝐢𝐧𝐡 𝟑 =𝟏𝟎.𝟎𝟐
Write in terms of 𝑒:
𝐜𝐨𝐬𝐞𝐜𝐡 𝟑 = 𝟏 𝐬𝐢𝐧𝐡 𝟑 = 𝟐 𝒆 𝟑 − 𝒆 −𝟑
Find the exact value of:
𝒕𝒂𝒏𝒉 𝒍𝒏 𝟒 = 𝒆 𝟐𝒍𝒏𝟒 −𝟏 𝒆 𝟐𝒍𝒏𝟒 +𝟏 = 𝒆 𝒍𝒏 𝟒 𝟐 −𝟏 𝒆 𝒍𝒏 𝟒 𝟐 +𝟏 = 𝟏𝟔−𝟏 𝟏𝟔+𝟏 = 𝟏𝟓 𝟏𝟕
Solve 𝑠𝑖𝑛ℎ 𝑥=5
𝒆 𝒙 − 𝒆 −𝒙 𝟐 =𝟓 ⇒ 𝒆 𝒙 − 𝒆 −𝒙 =𝟏𝟎 𝒆 𝟐𝒙 −𝟏𝟎 𝒆 𝒙 −𝟏=𝟎 𝒆 𝒙 = 𝟏𝟎± 𝟏𝟎𝟎+𝟒 𝟐 =𝟏𝟎.𝟎𝟗𝟗 𝒐𝒓 −𝟎.𝟎𝟗𝟗 𝒙=𝟐.𝟑𝟏 ?
Broculator Tip: Press the ‘hyp’ button. ? ? ?

7. Exercise 1
Use definitions of hyperbolic functions to find each answer, then check using inverse hyperbolic function on your calculator.
Solve to 2dp:
cosh 𝑥 =2 ⇒ 𝒙=𝟏.𝟑𝟐, 𝒙=−𝟏.𝟑𝟐 sinh 𝑥 =1 ⇒ 𝒙=𝟎.𝟖𝟖 tanh 𝑥 =− ⇒ 𝒙=−𝟎.𝟓𝟓 coth 𝑥 = ⇒ 𝒙=𝟎.𝟏𝟎 sech 𝑥 = ⇒ 𝒙=𝟐.𝟕𝟕, 𝒙=−𝟐.𝟕𝟕 2
Write in terms of 𝑒:
sinh 1 = 𝒆− 𝒆 −𝟏 𝟐 cosh 4 = 𝒆 𝟒 + 𝒆 −𝟒 𝟐 tanh = 𝒆−𝟏 𝒆+𝟏 sech −1 = 𝟐 𝒆 −𝟏 +𝒆
Find the exact values of:
sinh ln 2 = 𝟑 𝟒 cosh ( ln 3 )= 𝟓 𝟑 tanh ln 2 = 𝟑 𝟓 cosech ln 𝜋 = 𝟐𝝅 𝝅 𝟐 −𝟏 ? a ? b ? 4 ? c 5 ? ? ? d 6 ? 7 3 ? 8 ? a ? b ? c ? d

8. Click to sketch 𝑦= sinh 𝑥
Sketching hyperbolic functions 𝑦 𝑦
𝑦= 𝑒 𝑥
𝑦= 𝑒 −𝑥 𝑥 𝑥
𝑦=− 𝑒 𝑥
𝑦=− 𝑒 −𝑥 𝑦
𝑦= sinh 𝑥
Click to sketch 𝑦= sinh 𝑥 𝑥
As 𝑥→∞, 𝑒 −𝑥 →0 ∴ sinh 𝑥 → 1 2 𝑒 𝑥 ?
sinh 𝑥 = 𝑒 𝑥 − 𝑒 −𝑥 2 ?
We can see we have the average of 𝑒 𝑥 and − 𝑒 −𝑥

9. Click to sketch 𝑦= cosh 𝑥
Sketching hyperbolic functions 𝑦 𝑦
𝑦= 𝑒 𝑥
𝑦= 𝑒 −𝑥 𝑥 𝑥
𝑦=− 𝑒 𝑥
𝑦=− 𝑒 −𝑥 𝑦
𝑦= cosh 𝑥
Click to sketch 𝑦= cosh 𝑥 𝑥
As 𝑥→∞, cosh 𝑥 → 1 2 𝑒 𝑥 ?
cosh 𝑥 = 𝑒 𝑥 + 𝑒 −𝑥 2

10. Click to sketch 𝑦= tanh 𝑥
Sketching hyperbolic functions
tanh 𝑥 = sinh 𝑥 cosh 𝑥
To sketch 𝑦=tanh 𝑥 , consider the usual features when you sketch a graph.
When 𝑥=0, 𝒚= 𝐬𝐢𝐧𝐡 𝟎 𝐜𝐨𝐬𝐡 𝟎 = 𝟎 𝟏 =𝟎
As 𝑥→∞, 𝐭𝐚𝐧𝐡 𝒙 → 𝟏 𝟐 𝒆 𝒙 𝟏 𝟐 𝒆 𝒙 =𝟏
As 𝑥→−∞, 𝐭𝐚𝐧𝐡 𝒙 → − 𝟏 𝟐 𝒆 −𝒙 𝟏 𝟐 𝒆 −𝒙 =−𝟏 ? ? ? 𝑦
𝑦=1
Click to sketch 𝑦= tanh 𝑥 𝑥
𝑦=−1

11. Test Your Understanding
Sketch the graph of 𝑦= sech 𝑥 𝑥
𝑦= cosh 𝑥 1 ? ? 𝑥 𝑦 1
𝑦= sech 𝑥

12. Exercise 1B
On the same diagram, sketch 𝑦= cosh 2𝑥 and 𝑦=2 cosh 𝑥 . 1
b) Show that at the point of intersection, 𝑥= 1 2 ln
𝟐 𝒆 𝒙 + 𝒆 −𝒙 = 𝒆 𝒙 − 𝒆 −𝒙 𝟐 𝒆 𝒙 − 𝒆 −𝒙 𝒆 𝒙 + 𝒆 −𝒙 =𝟒 𝒆 𝟐𝒙 − 𝒆 −𝟐𝒙 =𝟒
Let 𝒚= 𝒆 𝟐𝒙 𝒚− 𝟏 𝒚 =𝟒 → 𝒚 𝟐 −𝟏=𝟒𝒚 𝒚 𝟐 −𝟒𝒚−𝟏=𝟎 𝒚= 𝟒± 𝟐𝟎 𝟐 =𝟐± 𝟓 𝒆 𝟐𝒙 =𝟐± 𝟓 𝒙= 𝟏 𝟐 𝐥𝐧 𝟐+ 𝟓 ? ?
𝑦=2 cosh 𝑥
𝑦= cosh 2𝑥 2
a) On the same diagram, sketch 𝑦= sech 𝑥 and 𝑦= sinh 𝑥 . ?
𝑦= sinh 𝑥
𝑦= sech 𝑥

13. Exercise 1B
Find the range of each hyperbolic function.
𝑓 𝑥 = sinh 𝑥 , 𝑥∈ℝ 𝒇 𝒙 ∈ℝ
𝑓 𝑥 = cosh 𝑥 , 𝑥∈ℝ 𝒇 𝒙 ≥𝟏
𝑓 𝑥 = tanh 𝑥 , 𝑥∈ℝ −𝟏<𝒇 𝒙 <𝟏
𝑓 𝑥 = sech 𝑥 , 𝑥∈ℝ 𝒇 𝒙 ≠𝟎, 𝒇 𝒙 ∈ℝ
𝑓 𝑥 = cosech 𝑥 , 𝑥∈ℝ 𝒇 𝒙 <−𝟏, 𝒇 𝒙 >𝟏
𝑓 𝑥 = coth 𝑥 , 𝑥∈ℝ 𝒇 𝒙 >𝟏
a) Sketch the graph of 𝑦=1+ coth 𝑥 , 𝑥∈ℝ,𝑥≠0 3 ? ? ? ? ? ? 4 ?
b) Write down the equation of the asymptotes to this curve.
𝒙=𝟎, 𝒚=𝟎, 𝒚=𝟐 ?

14. Exercise 1B ? ? 5 a) Sketch the graph of 𝑦=3 tanh 𝑥 , 𝑥∈ℝ
b) Write down the equation of the asymptotes to this curve.
𝒚=𝟑, 𝒚=−𝟑 ?

15. Hyperbolic Identities
From C2 we know that sin 2 𝑥 + cos 2 𝑥 =1.
Are there similar identities for hyperbolic functions?
Use the definitions of 𝑠𝑖𝑛ℎ and 𝑐𝑜𝑠ℎ to prove that… ?
𝐜𝐨𝐬𝐡 𝟐 𝒙 − 𝐬𝐢𝐧𝐡 𝟐 𝒙 =𝟏 ?
𝒔𝒆𝒄 𝒉 𝟐 𝒙=𝟏−𝒕𝒂𝒏 𝒉 𝟐 𝒙 ?
𝒄𝒐𝒔𝒆𝒄 𝒉 𝟐 𝒙=𝒄𝒐𝒕 𝒉 𝟐 𝒙−𝟏

16. Hyperbolic Identities
We can similar prove that:
Similar to sin⁡(𝐴+𝐵) identity.
𝒔𝒊𝒏𝒉 𝑨±𝑩 =𝒔𝒊𝒏𝒉 𝑨 𝒄𝒐𝒔𝒉 𝑩±𝒄𝒐𝒔𝒉 𝑨 𝒔𝒊𝒏𝒉 𝑩
𝒄𝒐𝒔𝒉 𝑨±𝑩 =𝒄𝒐𝒔𝒉 𝑨 𝒄𝒐𝒔𝒉 𝑩±𝒔𝒊𝒏𝒉 𝑨 𝒔𝒊𝒏𝒉 𝑩
However this is +, unlike in cos⁡(𝐴+𝐵)
Prove that:
𝒕𝒂𝒏𝒉 𝑨±𝑩 = 𝒔𝒊𝒏𝒉 𝒙 𝒄𝒐𝒔𝒉 𝒙 = 𝒕𝒂𝒏𝒉 𝑨+𝒕𝒂𝒏𝒉 𝑩 𝟏+𝒕𝒂𝒏𝒉 𝑨 𝒕𝒂𝒏𝒉 𝑩
Notice this is + rather than - .

17. Osborn’s Rule
We can get these identities from the normal sin/cos ones by:
Osborn’s Rule:
Replacing 𝑠𝑖𝑛→𝑠𝑖𝑛ℎ and 𝑐𝑜𝑠→𝑐𝑜𝑠ℎ
Negate any explicit or implied product of two sines.
sin 𝐴 sin 𝐵 →− sinh 𝐴 sinh 𝐵 tan 2 𝐴 →− tanh 2 𝐴 ? ?
Since tan 2 𝐴 = sin 2 𝐴 cos 2 𝐴 ?
𝑐𝑜𝑠2𝐴=2 cos 2 𝐴 −1 → 𝐜𝐨𝐬𝐡 𝟐𝑨 =𝟐𝐜𝐨𝐬 𝒉 𝟐 𝑨−𝟏 tan 𝐴−𝐵 = tan 𝐴 − tan 𝐵 1+ tan 𝐴 tan 𝐵 → 𝐭𝐚𝐧𝐡 𝑨 − 𝐭𝐚𝐧𝐡 𝑩 𝟏− 𝐭𝐚𝐧𝐡 𝑨 𝐭𝐚𝐧𝐡 𝑩 ?

18. Examples ? ? ? ? C2 one: If cos 𝑥 = 3 5 , find sin 𝑥 .
sin 2 𝑥 + cos 2 𝑥 =1 sin 𝑥 = 1− = 4 5 ?
If sinh 𝑥 = 3 4 , find the exact value of:
cosh 𝑥
tanh 𝑥
sinh 2𝑥
cosh 2 𝑥 − sinh 2 𝑥 =1 cosh 𝑥 = = tanh 𝑥 = sinh 𝑥 cosh 𝑥 = sinh 2𝑥 =2 sinh 𝑥 cosh 𝑥 = 15 8 ? ?

19. Exercise 1C ? ? ? ? ? Prove using the identities for 𝑠𝑖𝑛ℎ and 𝑐𝑜𝑠ℎ.
sinh 2𝐴 ≡2 sinh 𝐴 cosh 𝐴 cosh 𝐴−𝐵 ≡ cosh 𝐴 cosh 𝐵 − sinh 𝐴 sinh 𝐵 cosh 3𝐴 ≡4 cosh 3 𝐴 −3 cosh 𝐴 sinh 𝐴 − sinh 𝐵 ≡2 sinh 𝐴−𝐵 2 cosh 𝐴+𝐵 coth 𝐴 − tanh 𝐴 ≡2 cosech 2𝐴
Use Osborn’s rule to write down the corresponding hyperbolic identities:
sin 𝐴−𝐵 ≡ sin 𝐴 cos 𝐵 − cos 𝐴 sin 𝐵 sin 3𝐴 ≡3 sin 𝐴 −4 sin 3 𝐴 cos 𝐴 + cos 𝐵 ≡2 cos 𝐴+𝐵 2 cos 𝐴−𝐵 cos 2𝐴 ≡ 1− tan 2 𝐴 1+ tan 2 𝐴 cos 2𝐴 ≡ cos 4 𝐴 − sin 4 𝐴 1 2 3 4 5 6 ?
sinh 𝐴−𝐵 = sinh 𝐴 cosh 𝐴 − cosh 𝐴 sinh 𝐵 7 ?
sinh 3𝐴 =3 sinh 𝐴 +4 sinh 3 𝐴
cosℎ 𝐴 + cosh 𝐵 ≡2 cosh 𝐴+𝐵 2 cosh 𝐴−𝐵 2 ? 8
cosh 2𝐴 ≡ 1+ tanh 2 𝐴 1− tanℎ 2 𝐴 ? 9 ?
cosh 2𝐴 ≡ cosh 4 𝐴 − sinh 4 𝐴 10

20. Exercise 1C 11
Give that cosh 𝑥 =2, find the exact value of
sinh 𝑥 =± 𝟑
tanh 𝑥 =± 𝟑 𝟐
cosh 𝑥 =𝟕
Given that sinh 𝑥 =−1, find the exact value of
cosh 𝑥 = 𝟐
sinh 2𝑥 =−𝟐 𝟐
tanh 2𝑥 =− 𝟐 𝟐 𝟑 ? ? ? 12 ? ? ?

21. Inverse Hyperbolic Functions
As you might expect, each hyperbolic function has an inverse.
Note that lack of ‘c’. 𝑦
Click to sketch 𝑦= sinh 𝑥 𝑥
Click to sketch 𝑦= arsinh 𝑥
All of them:
𝒚=𝒂𝒓𝒔𝒊𝒏 𝒙, 𝒚=𝒂𝒓𝒄𝒐𝒔𝒉 𝒙, 𝒚=𝒂𝒓𝒕𝒂𝒏𝒉 𝒙 𝒚=𝒂𝒓𝒔𝒆𝒄𝒉 𝒙, 𝒚=𝒂𝒓𝒄𝒐𝒔𝒆𝒄𝒉 𝒙, 𝒚=𝒂𝒓𝒄𝒐𝒕𝒉 𝒙

22. Inverse Hyperbolic Functions
Why is there a problem when finding the inverse of 𝑓 𝑥 = cosh 𝑥 ? 𝑦
Recall from C3 that functions only have an inverse if they are one-to-one.
𝑓 𝑥 = cosh 𝑥 is many-to-one if the domain is unrestricted, which would become one-to-many. x
𝑦=𝑐𝑜𝑠ℎ 𝑥
𝑦=𝑎𝑟𝑐𝑜𝑠ℎ 𝑥 𝑥
If we restrict the domain to 𝑥≥0, then it becomes one-to-one, and we can reflect in 𝑦=𝑥 as before.

23. Inverse Hyperbolic Functions
Given that hyperbolic functions can be written in terms of 𝑒, naturally (obscure pun intended) inverse hyperbolic can be expressed in terms of 𝑙𝑛.
Prove that 𝑎𝑟𝑠𝑖𝑛ℎ 𝑥= ln 𝑥+ 𝑥 2 +1 ?
𝒚=𝒂𝒓𝒔𝒊𝒏𝒉 𝒙 𝒙= 𝐬𝐢𝐧𝐡 𝒚 𝒙= 𝒆 𝒚 − 𝒆 −𝒚 𝟐 𝒆 𝒚 − 𝒆 −𝒚 =𝟐𝒙 𝒆 𝟐𝒚 −𝟐𝒙 𝒆 𝒚 −𝟏=𝟎 𝒂=−𝟏, 𝒃=−𝟐𝒙, 𝒄=−𝟏 𝒆 𝒚 = 𝟐𝒙± 𝟒 𝒙 𝟐 +𝟒 𝟐 =𝒙± 𝒙 𝟐 +𝟏
However since 𝒙 𝟐 +𝟏 >𝒙, can only use positive case as 𝒆 𝒚 is positive.
𝒚= 𝐥𝐧 𝒙+ 𝒙 𝟐 +𝟏 𝒂𝒓𝒔𝒊𝒏𝒉 𝒙= 𝐥𝐧 𝒙+ 𝒙 𝟐 +𝟏 ? ?

24. Test Your Understanding
Prove that 𝑎𝑟𝑐𝑜𝑠ℎ 𝑥= ln 𝑥+ 𝑥 2 −1 , 𝑥≥1
Proof that 𝒍𝒏⁡(𝒙− 𝒙 𝟐 −𝟏 ) is negative: 𝑥− 𝑥 2 −1 𝑥+ 𝑥 2 −1 =1 𝑥− 𝑥 2 −1 = 1 𝑥+ 𝑥 2 −1
Taking logs of both sides:
ln 𝑥− 𝑥 2 −1 =− ln 𝑥+ 𝑥 2 −1
Since 𝑥≥1, 𝑥+ 𝑥 2 −1 ≥1, thus RHS is negative. ?
𝒚=𝒂𝒓𝒄𝒐𝒔𝒉 𝒙 𝒙= 𝐜𝐨𝐬𝐡 𝒚 𝒙= 𝒆 𝒚 + 𝒆 −𝒚 𝟐 𝒆 𝒚 + 𝒆 −𝒚 =𝟐𝒙 𝒆 𝟐𝒚 −𝟐𝒙 𝒆 𝒚 +𝟏=𝟎 𝒂=−𝟏, 𝒃=−𝟐𝒙, 𝒄=+𝟏 𝒆 𝒚 = 𝟐𝒙± 𝟒 𝒙 𝟐 −𝟒 𝟐 =𝒙± 𝒙 𝟐 −𝟏
However this time, both + and − cases are possible.
We can prove that ln⁡(𝑥− 𝑥 2 −1 ) gives a negative value.
Show >
But recall from the graph that we only include positive values of 𝑦 in the function to avoid it being one-to-many. Thus𝑎𝑟𝑐𝑜𝑠ℎ 𝑥= ln 𝑥+ 𝑥 2 −1 only.
𝑦=𝑎𝑟𝑐𝑜𝑠ℎ 𝑥

25. Summary
𝑎𝑟𝑠𝑖𝑛ℎ 𝑥= ln 𝑥+ 𝑥 𝑎𝑟𝑐𝑜𝑠ℎ 𝑥= ln 𝑥+ 𝑥 2 −1 , 𝑥≥1 𝑎𝑟𝑡𝑎𝑛ℎ 𝑥= 1 2 ln 1+𝑥 1−𝑥 , 𝑥 <1
Hyperbolic
Domain
Sketch
Inverse Hyperbolic
𝑦= sinh 𝑥
𝑥∈ℝ
𝑦=𝑎𝑟𝑠𝑖𝑛ℎ 𝑥
𝑦= cosh 𝑥
𝑥≥0
𝑦=𝑎𝑟𝑐𝑜𝑠ℎ 𝑥
𝑥≥1
𝑦= tanh 𝑥
𝑦=𝑎𝑟𝑡𝑎𝑛ℎ 𝑥
𝑥 <1
𝑦= sech 𝑥
𝑦=𝑎𝑟𝑠𝑒𝑐ℎ 𝑥
0<𝑥≤1
𝑦=𝑐𝑜𝑠𝑒𝑐ℎ 𝑥
𝑥≠0
𝑦=𝑎𝑟𝑐𝑜𝑠𝑒𝑐ℎ 𝑥
𝑦= coth 𝑥
𝑦= arcoth 𝑥
𝑥 >1 ? ? ? ? 1 1 ? 1 ? -1 ? ? 1 ? ? ? ? 1 -1

26. Exercise 1D 1 Sketch the graph of 𝑦=𝑎𝑟𝑡𝑎𝑛ℎ 𝑥, 𝑥 <1
Show that 𝑎𝑟𝑡𝑎𝑛ℎ 𝑥= 1 2 ln 1+𝑥 1−𝑥 , 𝑥 <1
Show that 𝑎𝑟𝑠𝑒𝑐ℎ 𝑥= ln − 𝑥 2 𝑥 , 0<𝑥≤1
Express as natural logarithms.
a) 𝑎𝑟𝑠𝑖𝑛ℎ b) 𝑎𝑟𝑐𝑜𝑠ℎ c) 𝑎𝑟𝑡𝑎𝑛ℎ 1 2
a) 𝑎𝑟𝑠𝑖𝑛ℎ b) 𝑎𝑟𝑐𝑜𝑠ℎ c) 𝑎𝑟𝑡𝑎𝑛ℎ 0.1
a) 𝑎𝑟𝑠𝑖𝑛ℎ − b) 𝑎𝑟𝑐𝑜𝑠ℎ c) 𝑎𝑟𝑡𝑎𝑛ℎ 1 3
Given that 𝑎𝑟𝑡𝑎𝑛ℎ 𝑥+𝑎𝑟𝑡𝑎𝑛ℎ 𝑦= ln 3 , show that 𝑦= 2𝑥−1 𝑥−2 2 3 4 5 6 7 8 9 10

27. Solving Equations
Either use hyperbolic identities or basic definitions of hyperbolic functions.
Solve for all real 𝑥
6 sinh 𝑥 −2 cosh 𝑥 =7
Solve for all real 𝑥
2 cosh 2 𝑥 − 5𝑠𝑖𝑛ℎ 𝑥 =5 ? ?
Using cosh 2 𝑥 −sin ℎ 2 𝑥=1
2 1+ sinh 2 𝑥 −5 sinh 𝑥 =5 2 sinh 𝑥 +1 sinh 𝑥 −3 =0 sinh 𝑥 =− 1 2 , sinh 𝑥 =3 𝑥=𝑎𝑟𝑠𝑖𝑛ℎ − 1 2 , 𝑥=𝑎𝑟𝑠𝑖𝑛ℎ 3 𝑥= ln − , 𝑥= ln
6 𝑒 𝑥 − 𝑒 −𝑥 2 −2 𝑒 𝑥 + 𝑒 −𝑥 2 =7 … 2 𝑒 𝑥 +1 𝑒 𝑥 −4 =0 𝑥= ln 4

28. Test Your Understanding? ?

29. Exercise 1E
Solve:
3 sinh 𝑥 +4 cosh 𝑥 = 𝒙=−𝐥𝐧 𝟕, 𝟎 7 sinh 𝑥 −5 cosh 𝑥 = 𝒙= 𝐥𝐧 𝟑 30 cosh 𝑥 =15+26 sinh 𝑥 𝒙= 𝐥𝐧 𝟕 𝟐 , 𝐥𝐧 𝟒 13 sinh 𝑥 −7 cosh 𝑥 +1= 𝒙= 𝐥𝐧 𝟓 𝟑 cosh 2𝑥 −5 sinh 𝑥 = 𝒙= 𝐥𝐧 −𝟑+ 𝟏𝟑 𝟐 , 𝐥𝐧 𝟒± 𝟏𝟕 2tan ℎ 2 𝑥+5 sech 𝑥 −4= 𝒙= 𝐥𝐧 𝟐± 𝟑 sinh 2 𝑥 −13 cosh 𝑥 +7= 𝒙= 𝐥𝐧 𝟒± 𝟏𝟓 sinh 2𝑥 −7 sinh 𝑥 = 𝒙=𝟎, 𝐥𝐧 𝟕±𝟑 𝟓 𝟐 4 cosh 𝑥 +13 𝑒 −𝑥 = 𝒙= 𝐥𝐧 𝟓 𝟐 , 𝐥𝐧 𝟑 2 tanh 𝑥 = cosh 𝑥 𝒙= 𝐥𝐧 𝟏+ 𝟐 ? 1 2 ? ? 3 ? 4 ? 5 6 ? ? 7 ? 8 ? 9 10 ? ?