# FP2 Hyperbolic Functions - Proof

Watch
Announcements
#1
Prove that cosh 2x = 2 cosh2 x - 1

My working:

cosh x = 1/2(ex + e-x)
cosh2 x = 1/4(e2x + 2e0 + e-2x)
2cosh2 x = 1/2(e2x + e-2x + 2)
2cosh2 x = 1/2(e2x + e-2x) + 1
2cosh2 x - 1 = 1/2(e2x + e-2x)
= cosh 2x

Is this enough? Are there any other ways to prove it?
0
14 years ago
#2
You could note that coshx = cos(ix), and thus the identity cos2a = 2cos2a - 1 holds.

Note though that sinhx = -isin(ix), and so cosh2x = cos2(ix) = 1 - sin2(ix) = 1 + (-i)2sin2(ix) = 1 + sinh2x. So it follows that cosh2x - sinh2x = 1, and also that cosh2x = cos2(ix) = cos2(ix) - sin2(ix) = cosh2x + sinh2x.

This can be verified using your method, of course:

We know that coshx = 1/2(ex + e-x), so cosh2x = 1/2(e2x + e-2x).
cosh2x = 1/4(e2x + e-2x + 2)
sinh2x = 1/4(e2x + e-2x - 2)
cosh2x + sinh2x = 1/2(e2x + e-2x) = cosh2x.

Bit irrelevant, but meh, useful. 0
14 years ago
#3
Looks like you did alright mate. I might have started with cosh(2x) = [e^(2x) + e^(-2x)]/2 and kept working.
0
#4
(Original post by Juwel)
Looks like you did alright mate. I might have started with cosh(2x) = [e^(2x) + e^(-2x)]/2 and kept working.
I tried this way but couldn't get actually get anywhere. Any ideas on how to start it off?

0
14 years ago
#5
(Original post by CalculusMan)
I tried this way but couldn't get actually get anywhere. Any ideas on how to start it off?

cosh(2x) = [e^(2x) + e^(-2x)]/2
= [(e^x)^2 + (e^-x)^2]/2
= [(e^x)^2 + (e^-x)^2 + 2 - 2]/2
= [(e^x + e^-x)^2 - 2]/2
= [(e^x + e^-x)^2]/2 - 1
= cosh^2 x - 1.
0
#6
You've just proved something which isn't true.

I've followed your working but can't seem to spot how it should be (2cosh^x - 1)
0
14 years ago
#7
cosh(2x) = [e^(2x) + e^(-2x)]/2
= [(e^x)^2 + (e^-x)^2]/2
= [(e^x)^2 + (e^-x)^2 + 2 - 2]/2
= [(e^x + e^-x)^2 - 2]/2
= [(e^x + e^-x)^2]/2 - 1
= cosh^2 x - 1.
cosh(2x) = [e^(2x) + e^(-2x)]/2
[e^(2x) + e^(-2x)]/2
= [(e^x)^2 + (e^-x)^2]/2
= [(e^x)^2 + (e^-x)^2 + 2 - 2]/2
= [(e^x + e^-x)^2 - 2]/2
= 2[(e^x + e^-x)^2]/4 - 1
= 2[(e^x + e^-x)/2]^2 - 1
= 2cosh^2 x - 1
0
14 years ago
#8
(Original post by Jan (Honza))
cosh(2x) = [e^(2x) + e^(-2x)]/2
[e^(2x) + e^(-2x)]/2
= [(e^x)^2 + (e^-x)^2]/2
= [(e^x)^2 + (e^-x)^2 + 2 - 2]/2
= [(e^x + e^-x)^2 - 2]/2
= 2[(e^x + e^-x)^2]/4 - 1
= 2[(e^x + e^-x)/2]^2 - 1
= 2cosh^2 x - 1
Oops. Looks like I've forgotten how to think. Thanks.
0
1 year ago
#9
Where is that -2 comming from the second = signe
0
1 month ago
#10
Senh^2 (x) -2cosh^2(2)=1
0
X

new posts Back
to top
Latest
My Feed

### Oops, nobody has postedin the last few hours.

Why not re-start the conversation?

see more

### See more of what you like onThe Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

### Poll

Join the discussion

#### Feeling behind at school/college? What is the best thing your teachers could to help you catch up?

Extra compulsory independent learning activities (eg, homework tasks) (12)
7.41%
Run extra compulsory lessons or workshops (27)
16.67%
Focus on making the normal lesson time with them as high quality as possible (26)
16.05%
Focus on making the normal learning resources as high quality/accessible as possible (24)
14.81%
Provide extra optional activities, lessons and/or workshops (45)
27.78%
Assess students, decide who needs extra support and focus on these students (28)
17.28%