So, for example, if θ = 2π, then θ = 2π ≠ cos-1(cos(2π)) = cos-1(1) = 0.

[EngnrMatt]

Homework Statement

cosθ = 1.316581

Homework Equations

cos^-1cosθ = θ // if and only if 0$\leqθ\leq1$

The Attempt at a Solution

This problem is actually the result of an attempt at a solution to solve an even larger problem (SAS triangle specifically). I need to know how to restrict the cosθ value in order to get θ. If it helps any, the original fraction was (49-196.457081)/-112

I know its got to be simple. I just have a terrible professor for math :P

 

[rock.freak667]

Well cosθ > 1 will yield no real solutions as -1≤cosθ≤1


Meaning that the maximum value of cosθ is 1 and the minimum value is -1.

 

[SammyS]

Homework Statement

cosθ = 1.316581

Homework Equations

cos^-1cosθ = θ // if and only if 0$\leqθ\leq1$

The Attempt at a Solution

This problem is actually the result of an attempt at a solution to solve an even larger problem (SAS triangle specifically). I need to know how to restrict the cosθ value in order to get θ. If it helps any, the original fraction was (49-196.457081)/-112

I know its got to be simple. I just have a terrible professor for math :P

As rock.freak667 point out, cos(θ) can never be greater than 1, nor less than -1.

Perhaps θ = 1.316581 and you need to find cos(∂).

As for your statement: cos^-1cosθ = θ // if and only if 0 ≤ θ ≤ 1 , that's incorrect.

The correct statement is

$\displaystyle \cos^{-1}(\cos(\theta))=\theta\ \ \text{ if and only if }\ \ 0\le\theta\le\pi\ .$​