Deriving expression as a function of another variable

[kenef]

I am attempting to understand how I can derive a new expression as a function of another variable in this existing expression. This is a college undergraduate course but I can't recall doing this much before, maybe in high school. I have taken all the way up to ODE's and PDE's.

The confusion that begins to arise is when attempting to manipulate this expression due to its trigonometric function.

With that being said, if one has an equation such that:

B(r) = Bsin(A f r + θ)

How can one solve to express this as a function of let's say f?
Any insight towards mathematical properties would be greatly appreciated!

 

[mfb]

What does "Bsin(A f r + θ)" mean? Is B a constant here? On the left side it appears to be a function.

You'll need inverse trigonometric functions.

 

[kenef]

What does "Bsin(A f r + θ)" mean? Is B a constant here? On the left side it appears to be a function.

You'll need inverse trigonometric functions.

Yes B is a constant, let's say the original amount.

I was thinking that perhaps one might need to use Laplace transform

 

[jtbell]

If $x = \sin(y)$, then $y = \sin^{-1}(x)$. Or to use an alternative notation, $y = \arcsin(x)$. But surely you've seen this already, if you've studied ODEs and PDEs.

 

[Mark44]

B(r) = Bsin(A f r + θ)

I was thinking that perhaps one might need to use Laplace transform

Laplace transforms? No, this is much simpler than that.

Let's rewrite the equation above as g(r) = Bsin(Afr + θ) so that B doesn't have two different roles.
1. Divide both sides by B
2. Take the inverse sine of both sides
3. Etc.

 

[Mark44]

Thread moved as it doesn't appear to have anything to do directly with Calculus...

 

[kenef]

Thread moved as it doesn't appear to have anything to do directly with Calculus...

Thank you, it was originally under General Math, and someone moved it to Calculus. Its not really related to "Calculus" as you have said. I will take your suggestion for the solution, thank you!