Solving trigonometric equations involving arctan?

**niggawut** · October 1st 2011, 12:33 AM

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:

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

a, b and C are constants.

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

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

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

**niggawut** · October 1st 2011, 12:43 AM

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

**Siron** · October 1st 2011, 02:13 AM

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

**niggawut** · October 1st 2011, 02:22 AM

thanks.

**niggawut** · October 1st 2011, 02:28 AM

Originally Posted by Siron

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

So x is simply any angle?

Also, how do you input math notation like that?

**Siron** · October 1st 2011, 02:42 AM

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

**niggawut** · October 1st 2011, 04:36 AM

Ok so i used the theorem and ended up with:

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

Yeah I learned how to use the notation.

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

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

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

**niggawut** · October 1st 2011, 05:16 AM

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

**niggawut** · October 2nd 2011, 06:16 AM

Originally Posted by niggawut

I tried resolving it into a quadratic equation in terms of $sin(\theta)$ , 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.

**niggawut** · October 3rd 2011, 04:29 AM

perhaps my co-effs are wrong.

Thread: Solving trigonometric equations involving arctan?

LinkBack

Thread Tools

Display

Solving trigonometric equations involving arctan?

Re: Solving trigonometric equations involving arctan?

Re: Solving trigonometric equations involving arctan?

Re: Solving trigonometric equations involving arctan?

Re: Solving trigonometric equations involving arctan?

Re: Solving trigonometric equations involving arctan?

Re: Solving trigonometric equations involving arctan?

Re: Solving trigonometric equations involving arctan?

Re: Solving trigonometric equations involving arctan?

Re: Solving trigonometric equations involving arctan?

Similar Math Help Forum Discussions

Solving for x in equations involving logs

Solving Trigonometric equations 2

Solving equations involving logs

solving equations involving inequalities

Solving Trigonometric Equations

Search Tags