marcsaccount
Badges: 1
Rep:
?
#1
Report Thread starter 6 years ago
#1
Hi,

I've got a problem understanding inverse trig functions. Frequently during c2, I took the inverse of a trig value, and my understanding was that  sin^1 y=x and I thought of this as if we know the sin value (y), if we took the inverse of it, this would give us the angle, x. So, undoing the sin of y, gives x. This makes sense to me.

Now at c3, they say it is  sin^1 x=y. This doesn't make sense to me because the y value on a trig graph has the sin function applied to it, so any y value, say 0.5 is a tig value. I don't understand!
0
reply
Old_Simon
Badges: 12
Rep:
?
#2
Report 6 years ago
#2
The angle of x got nothing to do with the inverse.
0
reply
Kvothe the Arcane
Badges: 20
Rep:
?
#3
Report 6 years ago
#3
(Original post by marcsaccount)
Hi,

I've got a problem understanding inverse trig functions. Frequently during c2, I took the inverse of a trig value, and my understanding was that  sin^1 y=x and I thought of this as if we know the sin value (y), if we took the inverse of it, this would give us the angle, x. So, undoing the sin of y, gives x. This makes sense to me.

Now at c3, they say it is  sin^1 x=y. This doesn't make sense to me because the y value on a trig graph has the sin function applied to it, so any y value, say 0.5 is a tig value. I don't understand!
Can you write your question differently please?
I'm afraid I don't understand but I'm willing to help.
0
reply
davros
  • Study Helper
Badges: 16
Rep:
?
#4
Report 6 years ago
#4
(Original post by marcsaccount)
Hi,

I've got a problem understanding inverse trig functions. Frequently during c2, I took the inverse of a trig value, and my understanding was that  sin^1 y=x and I thought of this as if we know the sin value (y), if we took the inverse of it, this would give us the angle, x. So, undoing the sin of y, gives x. This makes sense to me.

Now at c3, they say it is  sin^1 x=y. This doesn't make sense to me because the y value on a trig graph has the sin function applied to it, so any y value, say 0.5 is a tig value. I don't understand!
I'm not quite sure what your confusion is.

You can define a functional relationship between any two variables - e.g. you can say y = x^2 or x = y^2

As long as you know how to invert a (1-1) function defined between a particular domain and range, that is all you have to worry about.

So y = sin^{-1} x \implies x = sin y and x = sin^{-1} y \implies y = sin x

You have to be careful going in the opposite direction because sin (and cos) are many-to-one so you have to use the principal value when inverting.
0
reply
marcsaccount
Badges: 1
Rep:
?
#5
Report Thread starter 6 years ago
#5
Thanks for everybody's help and replies.

(Original post by davros)
y = sin^{-1} x \implies x = sin y and x = sin^{-1} y \implies y = sin x
OK, I guess my confusion is because  y=sinx So the inverse is  x=sin^{-1}y Say if x=45, sinx=y, y would = 0.707 (approx), then it would be the inverse of y that would = x so  x=sin^{-1}y, not  y=sin^{-1}x
0
reply
davros
  • Study Helper
Badges: 16
Rep:
?
#6
Report 6 years ago
#6
(Original post by marcsaccount)
Thanks for everybody's help and replies.



OK, I guess my confusion is because  y=sinx So the inverse is  x=sin^{-1}y Say if x=45, sinx=y, y would = 0.707 (approx), then it would be the inverse of y that would = x so  x=sin^{-1}y, not  y=sin^{-1}x
Basically, yes.

You have to be careful because multiple angles can give the same sine value, but it's purely conventional that you normally plot y against x and not the other way round!
1
reply
marcsaccount
Badges: 1
Rep:
?
#7
Report Thread starter 6 years ago
#7
Thanks for your help Davros
0
reply
X

Quick Reply

Attached files
Write a reply...
Reply
new posts
Back
to top
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

Current uni students - are you thinking of dropping out of university?

Yes, I'm seriously considering dropping out (139)
14.33%
I'm not sure (42)
4.33%
No, I'm going to stick it out for now (289)
29.79%
I have already dropped out (25)
2.58%
I'm not a current university student (475)
48.97%

Watched Threads

View All