# Help with entering an equation into a calculator

1. May 4, 2008

### Etherian

First I should mention I'm actually going to use this in actionscript (a programming language), but as most people here probably don't know actionscript, I thought a graphing calculator would be a good analogue. I am also using an actual calculator, though, and will from here on regard the question as if it was for the calculator.

1. The problem statement, all variables and given/known data
I need to enter this equation into a calculator with $$\theta$$ isolated:
$$S=\frac{3cos^{2}\theta-1}{2}$$
The equation gives how many degrees off a liquid crystal will be from the local director ($$\theta$$) depending on on the molecule's order parameter (S). The lower the order parameter the farther off the director the molecule can be. 0 is no order, 1 is perfect order.

2. Relevant equations
I need to type this into a calculator where S is the input and $$\theta$$ is the output. I'm only in 8th grade and we haven't worked with these kind of equations yet, as simple as it is, so I'm not sure how to type it, let alone isolate $$\theta$$.

3. The attempt at a solution
I haven't tried isolating $$\theta$$ yet as can't even type the original it in correctly.
What I have so far is S=(3[cos($$\theta$$)^2]-1)/2.
This looks correct to me, but when I enter a number for S it gives a result I don't believe could be correct. I have entered it two or three times and, unfortunately, I seemed to have entered it differently both times, giving me an even less realistic output.

I need a response fairly quickly, since the project is due tomorrow. Thanks in advance to anyone that helps.

Last edited: May 4, 2008
2. May 4, 2008

### HallsofIvy

Staff Emeritus
If $S=(3[cos^2(\theta)]-1)/2[/tex]. then [itex]2S= 3 cos^2(\theta)- 1$, $2S+ 1= 3 cos^2(\theta)$, $(2S+1)/3= cos^2(\theta)$, $\sqrt{(2S+1)/3}= cos(\theta)$ and, finally, $\theta= cos^{-1}(\sqrt{(2S+1)/3})$.

That "cos-1" is, of course, the inverse- or arc-cosine, not the reciprocal.

3. May 4, 2008

### Etherian

Thanks for responding so quickly (in fact faster than I was able to edit my own post). The equation still won't work, but I believe it is that I misunderstood the equation, than it was incorrect or someone did something wrong. Thanks again for the help.