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Finding x in an inverse trigonometric equation

  • Thread starter acen_gr
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  • #1
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Homework Statement


Given y = 2/3(arctan [itex]\frac{x}{4}[/itex]), find x.

Homework Equations


Inverse trigonometric identities for tan. (I guess)


The Attempt at a Solution


I never attempted because it seems impossible for me. You think this is possible? Thank you!
 

Answers and Replies

  • #2
Mentallic
Homework Helper
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If y = arctan(x), what is x?

And by "what is x" or "find x" we mean to make x the subject, so we'll have something like x = ... (or if you understand how functions work, x will be a function of y which can also be written as x=f(y) ).
 
  • #3
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If y = arctan(x), what is x?

And by "what is x" or "find x" we mean to make x the subject, so we'll have something like x = ... (or if you understand how functions work, x will be a function of y which can also be written as x=f(y) ).
Yes. That's why I'm so confused with this. Is it just like "what must be the value of x to be y"?. At first, I really don't get this. But I think we should set y = 0 (like the way of how x-intercepts are found). Is it possible to find x that way? Or is this problem again makes no sense?

EDIT: Or do I make no sense with this reply?
 
  • #4
Mentallic
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Yes. That's why I'm so confused with this. Is it just like "what must be the value of x to be y"?. At first, I really don't get this. But I think we should set y = 0 (like the way of how x-intercepts are found). Is it possible to find x that way? Or is this problem again makes no sense?

EDIT: Or do I make no sense with this reply?
If y=3x, then x=y/3

If [itex]y=x^2[/itex], then [itex]x=\pm\sqrt{y}[/itex]

If [itex]y=arctan(x)[/itex] then [itex]x=...[/itex]?

In other words, what do you do to arctan(x) to make it go back to x again? What we did to x2 is we took the square root (and we have to take the square root of both sides). What we did to 3x is we divided it by 3 (and we divided both sides). This operation is called taking the inverse.

The inverse of squaring, is square-rooting. The inverse of multiplying is dividing. The inverse of sin is arcsin (or sin-1) and the inverse of arcsin is sin. So what is the inverse of arctan?
 
  • #5
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If y=3x, then x=y/3

If [itex]y=x^2[/itex], then [itex]x=\pm\sqrt{y}[/itex]

If [itex]y=arctan(x)[/itex] then [itex]x=...[/itex]?

In other words, what do you do to arctan(x) to make it go back to x again? What we did to x2 is we took the square root (and we have to take the square root of both sides). What we did to 3x is we divided it by 3 (and we divided both sides). This operation is called taking the inverse.

The inverse of squaring, is square-rooting. The inverse of multiplying is dividing. The inverse of sin is arcsin (or sin-1) and the inverse of arcsin is sin. So what is the inverse of arctan?
I know! I know! Should we take the tangent? of both sides?
 
  • #6
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[tex]y = \frac{2}{3}(arctan\frac{x}{4})[/tex]

leave arctan alone:
[tex]\frac{3}{2}y = arctan\frac{x}{4}[/tex]

take tangent of both sides:
[tex]tan\frac{3}{2}y = \frac{x}{4}[/tex]

I should stop right here. Am I doing it right?

EDIT: To continue,

[tex]x = 4tan\frac{3}{2}y[/tex]
 
  • #7
Mentallic
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I know! I know! Should we take the tangent? of both sides?
Yes! If [tex]y=\arctan(x)[/tex] then to simplify it we take the tangent of both sides: [tex]\tan(y)=\tan(\arctan(x))[/tex]

And the tan and arctan cancel each other, so we end up with [tex]x=\tan(y)[/tex]

Now, for your problem, you don't want to take the tangent straight away, because your question isn't in the form [tex]y=\arctan(x)[/tex] you instead have a constant multiplied in front of the arctan, so you should get that to the other end first before taking the tangent.
 
  • #8
Mentallic
Homework Helper
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[tex]y = \frac{2}{3}(arctan\frac{x}{4})[/tex]

leave arctan alone:
[tex]\frac{3}{2}y = arctan\frac{x}{4}[/tex]

take tangent of both sides:
[tex]tan\frac{3}{2}y = \frac{x}{4}[/tex]

I should stop right here. Am I doing it right?

EDIT: To continue,

[tex]x = 4tan\frac{3}{2}y[/tex]
Yep! :smile:

Now just be sure to put brackets around 3y/2 because you can easily mistake it for something else like tan(3/2)*y which would be wrong.
 
  • #9
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Then the answer should be

[tex]x = 4tan(\frac{3y}{2})[/tex]

YES! Finally I made something right in my life again! Especially in math. Thank you! :)

@Offtopic, you really are a great teacher in math. Do you teach? Are you a professor or the like?
 
  • #10
Mentallic
Homework Helper
3,798
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Then the answer should be

[tex]x = 4tan(\frac{3y}{2})[/tex]

YES! Finally I made something right in my life again! Especially in math. Thank you! :)
No problem dude! And I'm glad to see that you're making progress.

@Offtopic, you really are a great teacher in math. Do you teach? Are you a professor or the like?
Cheers :smile: Nah I'm just a math student myself. But I've been on this forum for like 3 or 4 years now, so I've learnt a thing or two about how to teach others while not breaking the forum rules by giving out full solutions and the like.
 
  • #11
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Cheers :smile: Nah I'm just a math student myself.
Really? How come you seem to know a lot in math? Have you just memorized everything about math?

EDIT: Wee I'm so encouraged to work hard in math...knowing that even a math student can also be as good as teachers, and sometimes BETTER. I wish I could be like you too. You're a student like me yet you are like a math teacher! How did you become a homework helper?
 
Last edited:
  • #12
Mentallic
Homework Helper
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Really? How come you seem to know a lot in math? Have you just memorized everything about math?
Far from it. I haven't forgotten the important formulae that get me through the day, but to solve equations, puzzles, and basically just applying your knowledge of maths becomes a lot easier when you have a firm grasp of your algebra, trig and calculus.

I mean, if you looked at some early high school problems, even though you might have never answered that particular problem before, you would probably have an easy time answering it, right? Well, it's the same situation here.
 
  • #13
63
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Far from it. I haven't forgotten the important formulae that get me through the day, but to solve equations, puzzles, and basically just applying your knowledge of maths becomes a lot easier when you have a firm grasp of your algebra, trig and calculus.

I mean, if you looked at some early high school problems, even though you might have never answered that particular problem before, you would probably have an easy time answering it, right? Well, it's the same situation here.
You mean, you learned to solve anything because you have already encountered them? Or the opposite?
So, I should have studied a lot in algebra, trigo and calculus?
 

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