# Thread: Solving trigonometric equations involving arctan?

1. ## Solving trigonometric equations involving arctan?

I'm having difficulty solving the following equation which helps locate the angles in young's double slit experiment. Currently I'm using a graphical method.

The equation is:

$\displaystyle C^2 - b^2 - a^2$ = $\displaystyle \frac{a}{\tan^2(\theta)} + \frac{2aC}{\sin(\theta)} - \frac{2ba}{\tan(\theta)} - \frac{a^2}{\sin^2(\theta)}$

a, b and C are constants.

I tried expressing $\displaystyle {\tan(\theta)}$ in terms of $\displaystyle {\sin(\theta)}$.

However,$\displaystyle {\tan(\theta)}$ = ±$\displaystyle \frac{\sin(\theta)}{\sqrt{\1-sin^2(\theta)}}$

And I'm not sure how to deal with the ±, given that I would like to make θ the subject, where 0 < θ < $\displaystyle \pi$.

2. ## Re: Solving trigonometric equations involving arctan?

how do you express arctan in terms of basic trigonometric functions?

3. ## Re: Solving trigonometric equations involving arctan?

You can use the fact that $\displaystyle \sin\left[\arctan(x)\right]=\frac{x}{\sqrt{1+x^2}}$

thanks.

5. ## Re: Solving trigonometric equations involving arctan?

Originally Posted by Siron
You can use the fact that $\displaystyle \sin\left[\arctan(x)\right]=\frac{x}{\sqrt{1+x^2}}$
So x is simply any angle?

Also, how do you input math notation like that?

6. ## Re: Solving trigonometric equations involving arctan?

$\displaystyle x$ isn't an angle, it's a real number, for example $\displaystyle \arctan(1)=\frac{\pi}{4}+k\pi$ (Have you ever worked with cyclometric functions before?)
In this case $\displaystyle x=\frac{a}{\left|b-\left(\frac{a}{\tan(\theta)}\right)\right|}$ so $\displaystyle \sin[\arctan(x)]=...$?

7. ## Re: Solving trigonometric equations involving arctan?

Ok so i used the theorem and ended up with:

$\displaystyle C^2 - b^2 - a^2$ = $\displaystyle \frac{a}{\tan^2(\theta)} + \frac{2aC}{\sin(\theta)} - \frac{2ba}{\tan(\theta)} - \frac{a^2}{\sin^2(\theta)}$

Yeah I learned how to use the notation.

I tried expressing $\displaystyle {\tan(\theta)}$ in terms of $\displaystyle {\sin(\theta)}$.

However,$\displaystyle {\tan(\theta)}$ = ±$\displaystyle \frac{\sin(\theta)}{\sqrt{\1-sin^2(\theta)}}$

And I'm not sure how to deal with the ±, given that I would like to make θ the subject, where 0 < θ < $\displaystyle \pi$.

8. ## Re: Solving trigonometric equations involving arctan?

I tried resolving it into a quadratic equation in terms of $\displaystyle sin(\theta)$ , but it always gives me answers greater than 1.

9. ## Re: Solving trigonometric equations involving arctan?

Originally Posted by niggawut
I tried resolving it into a quadratic equation in terms of $\displaystyle sin(\theta)$ , but it always gives me answers greater than 1.
I squared both sides in order to form the quadratic equation but it doesnt seem to be working.

10. ## Re: Solving trigonometric equations involving arctan?

perhaps my co-effs are wrong.