Sign in to follow this  
CDProp

Solving a parameterized equation where the parameter is passed to a trig function...

Recommended Posts

I'm not sure if I'm using the correct terminology so I'll explain it this way. If I have the following function: f(x) = 3x And I want to know what value of x will return 15, I simply do this: 15 = 3x and solve for x, which is 5 in this case. However, what if the parameter is passed to a trig function like this: f(x) = cos(x) if I want to know what value of x that will return some value, is it possible? Intuitively, it seems impossible for two reasons: a) There are infinite solutions. If the output I'm solving for is 1.0, then there will be a solution for every other multiple of pi. b) As far as I know, before you solve an equation like this, you need to make sure that all functions within the equations are expanded. So, if I have the following equation: f(x) = g(x) + 3 and I want to know what the value of x would be if the value of the function were 6: 6 = g(x) + 3 I couldn't possible do this unless I knew what g(x) was. The best I could do is: 3 = g(x) And if g(x) = x + 2, then I could do this: 6 = x + 2 + 3 or x = 1 And since I don't know how to expand a trig function, I couldn't possibly get it in a form that's solvable. But I would really, really like to because I spent all day deriving a parameterized equation to solve a specific problem and it has like 8 trig functions in it, lol.

Share this post


Link to post
Share on other sites
I don't think it's explicitly possible. There might be some specific cases where you can, but not in general. Instead you have to either:

1. Use an iterative root finding method. -or-
2. Use a close approximation to the actual function.

Depending on your domain, I'd recommend using #2.

Share this post


Link to post
Share on other sites
You can use the inverse trig functions if you just have 1 trig function, but if you have several in one equation you'll probably have to find the answer numerically.

If you know f(x) and you want x from f(x)=cos(x) then x =acos(x). However, that only returns the value in the range near 0. If you want more values you'll have to loop through and add multiples of pi. If you post your derived equations I might be able to help you solve them specifically.

Share this post


Link to post
Share on other sites

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!

Register a new account

Sign in

Already have an account? Sign in here.

Sign In Now

Sign in to follow this