Differentiate y = arccosec(x) Watch

Zii
Badges: 13
Rep:
?
#1
Report Thread starter 10 years ago
#1
I get - \frac{1}{x\sqrt{x^2-1}}. Can someone else try it and tell me if I'm right?
0
reply
Speleo
Badges: 7
Rep:
?
#2
Report 10 years ago
#2
yep
0
reply
generalebriety
Badges: 14
Rep:
?
#3
Report 10 years ago
#3
Yep, same as me.

Edit: bah, snap.
0
reply
Zii
Badges: 13
Rep:
?
#4
Report Thread starter 10 years ago
#4
Thanks, guys. I'm trying to show for myself why \mathrm{arcsin}\frac{1}{x} = \mathrm{arccosec}x by differentiating them both and integrating them with limits to show that they are the same. Is this a good way of doing it?
0
reply
Speleo
Badges: 7
Rep:
?
#5
Report 10 years ago
#5
No not really. Do it anyway though, it might be instructive.

arcsin(1/x) = y
siny = 1/x
cosecy = x
y = arccosecx

EDIT: actually, you need to justify the first and last steps by looking at domains and ranges of the functions.
0
reply
madima
Badges: 0
Rep:
?
#6
Report 10 years ago
#6
is this stuff done in further maths or C3?

EDIT: nevermind. just did it using C3 knowledge.
0
reply
Zii
Badges: 13
Rep:
?
#7
Report Thread starter 10 years ago
#7
(Original post by Speleo)
No not really. Do it anyway though, it might be instructive.

arcsin(1/x) = y
siny = 1/x
cosecy = x
y = arccosecx

EDIT: actually, you need to justify the first and last steps by looking at domains and ranges of the functions.
Yeah you're totally right, it's a stupid way of doing it. Mind, after showing \mathrm{arcsin}\frac{1}{x} differentiates to give - \frac{1}{x\sqrt{x^2-1}}...

- \displaystyle\int^1_{-1} \frac{1}{x\sqrt{x^2-1}} \, \mathrm{d}x = \mathrm{arccosec}(1) - \mathrm{arccosec}(-1) = - \left(\frac{\pi}{2} - (-\frac{\pi}{2})\right) = -\pi

= \mathrm{arcsin}\frac{1}{1} - \mathrm{arcsin}\frac{1}{-1} = - \left(\frac{\pi}{2} - (-\frac{\pi}{2})\right) = - \pi

\rightarrow \mathrm{arccosec}x = \mathrm{arcsin}\frac{1}{x}
0
reply
DavyS
Badges: 0
Rep:
?
#8
Report 10 years ago
#8
(Original post by madima)
is this stuff done in further maths or C3?

EDIT: nevermind. just did it using C3 knowledge.
Although the actual content required to do it is in C3, its formally introduced (atleast on our exam board) in FP2.
0
reply
Speleo
Badges: 7
Rep:
?
#9
Report 10 years ago
#9
Actually that's quite a clever way to do it...

For x>0, since the functions agree at 0 you can take the integral of -1/tsqrt(t^2-1) dt between 0 and x to show that arccosecx = arcsin(1/x) and similarly for x<0.
0
reply
Cities
Badges: 14
Rep:
?
#10
Report 10 years ago
#10
(Original post by Speleo)
Actually that's quite a clever way to do it...

For x>0, since the functions agree at 0 you can take the integral of -1/tsqrt(t^2-1) dt between 0 and x to show that arccosecx = arcsin(1/x) and similarly for x<0.
Ah I was just going to say that; using a general case rather than the specific one Zii used.
0
reply
Zii
Badges: 13
Rep:
?
#11
Report Thread starter 10 years ago
#11
Edit: Does the fact that there's a discontinuity at zero not generate a big problem in my working?
0
reply
Speleo
Badges: 7
Rep:
?
#12
Report 10 years ago
#12
I didn't actually check that 0 worked, makes no difference though, replace (literally) every instance of '0' in my last post with '1'.
0
reply
Zii
Badges: 13
Rep:
?
#13
Report Thread starter 10 years ago
#13
I suppose I could have just integrated between 1 and 2 or something. It's just basically to show that the two potentially different constants generated are in fact equal (i.e. the answers are arcsin(1/x) + C or arccosec(x) + K - - - but C and K are in fact equal). C and K are shown to be equal by integrating within limits.
0
reply
Speleo
Badges: 7
Rep:
?
#14
Report 10 years ago
#14
Oh I see. It seems more solid to me to show that arcsin(1/x) = the definite integral dt + pi/2 = arccosec(x) for all x in the appropriate ranges but it doesn't really matter I suppose.
0
reply
Zii
Badges: 13
Rep:
?
#15
Report Thread starter 10 years ago
#15
I don't know what you mean "the definite integral + pi/2" ?
0
reply
Speleo
Badges: 7
Rep:
?
#16
Report 10 years ago
#16
[[the integral of -1/t*sqrt(t^2-1) dt between 1 and x]] + pi/2
0
reply
Zii
Badges: 13
Rep:
?
#17
Report Thread starter 10 years ago
#17
(Original post by Speleo)
[[the integral of -1/t*sqrt(t^2-1) dt between 1 and x]] + pi/2
Why the + \frac{\pi}{2}?

Edit: never mind, I get why now. Just scribbled it on a piece of paper .
0
reply
Speleo
Badges: 7
Rep:
?
#18
Report 10 years ago
#18
[[the integral of -1/t*sqrt(t^2-1) dt between 1 and x]] = arcsin(1/x) - arcsin(1/1) = arcsin(1/x) - pi/2
[[the integral of -1/t*sqrt(t^2-1) dt between 1 and x]] = arccosec(x) - arccosex(1) = arccosec(x) - pi/2

=>
arcsin(1/x) = [[the integral of -1/t*sqrt(t^2-1) dt between 1 and x]] + pi/2 = arccosec(x)

Choosing to use a definite integral is mostly personal preference, indefinite integrals are messier.
0
reply
Cities
Badges: 14
Rep:
?
#19
Report 10 years ago
#19
(Original post by Zii)
Why the + \frac{\pi}{2}?

Edit: never mind, I get why now. Just scribbled it on a piece of paper .
Out of curiosity, are you just figuring this out for fun, or is it for some kind of assignment? :p:
0
reply
Zii
Badges: 13
Rep:
?
#20
Report Thread starter 10 years ago
#20
(Original post by n1r4v)
Out of curiosity, are you just figuring this out for fun, or is it for some kind of assignment? :p:
For fun.

(Original post by Speleo)
[[the integral of -1/t*sqrt(t^2-1) dt between 1 and x]] = arcsin(1/x) - arcsin(1/1) = arcsin(1/x) - pi/2
[[the integral of -1/t*sqrt(t^2-1) dt between 1 and x]] = arccosec(x) - arccosex(1) = arccosec(x) - pi/2

=>
arcsin(1/x) = [[the integral of -1/t*sqrt(t^2-1) dt between 1 and x]] + pi/2 = arccosec(x)

Choosing to use a definite integral is mostly personal preference, indefinite integrals are messier.
Yeah I've never actually done that before, wow!

I just tried it with:

\displaystyle\int^x_0 \mathrm{sin}t\mathrm{cos}t \, \mathrm{d}t

= \frac{1}{2}\mathrm{sin}^2x - \frac{1}{2}\mathrm{sin}^2(0)

OR

= \frac{1}{4} - \frac{1}{4}\mathrm{cos}2x

(after doing the necessary work...first time was a simple "derivative sitting beside it job" and the second I rewrote \mathrm{sin}t\mathrm{cos}t as \frac{1}{2}\mathrm{sin}2t)

Which means:

\frac{1}{2}\mathrm{sin}^2x = \frac{1}{4} - \frac{1}{4}\mathrm{cos}2x

2\mathrm{sin}^2x = 1 - \mathrm{cos}2x

\mathrm{cos}2x = \mathrm{cos}^2x-\mathrm{sin}^2x

Cool stuff.
0
reply
X

Quick Reply

Attached files
Write a reply...
Reply
new posts
Latest
My Feed

See more of what you like on
The Student Room

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

Personalise

University open days

  • Cardiff Metropolitan University
    Undergraduate Open Day - Llandaff Campus Undergraduate
    Sat, 27 Apr '19
  • University of East Anglia
    Could you inspire the next generation? Find out more about becoming a Primary teacher with UEA… Postgraduate
    Sat, 27 Apr '19
  • Anglia Ruskin University
    Health, Education, Medicine and Social Care; Arts, Humanities and Social Sciences; Business and Law; Science and Engineering Undergraduate
    Sat, 27 Apr '19

Have you registered to vote?

Yes! (546)
37.73%
No - but I will (113)
7.81%
No - I don't want to (100)
6.91%
No - I can't vote (<18, not in UK, etc) (688)
47.55%

Watched Threads

View All