# Trigonometric identity

1. Aug 18, 2011

### Ry122

In the following question I figure that i need to prove that h holds true with the trigonometric identity subbed into the denominator.
I'm not sure how to simplify the equation any further after that though.
Can someone provide any insight?

http://img717.imageshack.us/img717/2661/parallax.jpg [Broken]

Last edited by a moderator: May 5, 2017
2. Aug 18, 2011

### micromass

Staff Emeritus
Re: parllax

Forget about the given formula for a moment. Try to find a good expression for the height h.

You could start by drawing the perpendicular on the baseline through the other point of the triangle. This will divide the baseline in section c' and c''.

Now, can you express h in function of c' and alpha?
And can you express h in function of c'' and beta?

3. Aug 20, 2011

### Ry122

Re: parllax

tanb=h/c'
tana=h/c

Substituting I end up with tana/tanb = c'/c

But what does this show me?

4. Aug 20, 2011

### micromass

Staff Emeritus
Re: parllax

Indeed, so

$$c^\prime=\frac{h}{\tan{\beta}}~\text{and}~c=\frac{h}{\tan{\alpha}}$$

Now we also know that $c=c^\prime+c^{\prime \prime}$

So what do you get if you plug in these values of c' and c''??

5. Aug 21, 2011

### Ry122

Re: parllax

Using the equations for c, c' and c'', this is as close as I got to the original formula:

h = c*sin(b)/cos(b)
+ c*sin(a)/cos(a)

What further steps do I need to take?

6. Aug 21, 2011

### dynamicsolo

Re: parllax

$c = h (\frac{cos\alpha}{sin\alpha} + \frac{cos\beta}{sin\beta} )$ ,

which you would then solve for h . (It will help to add the ratios first.)

Last edited: Aug 21, 2011
7. Aug 21, 2011

### Ry122

Re: parllax

that's exactly what I have, but with c as the subject.
Just wondering how to get it into a similar form to the original formula.

8. Aug 21, 2011

### dynamicsolo

Re: parllax

What you have is not equivalent to c = c' + c'' . (It is not correct to add those terms.)

Using the terms micromass has in post #4 , you have h = c' tan(beta) and, separately,
h = c'' tan(alpha) [he has a typo] . (And I had to edit my typo in post #6.)

The equation in post #6 is what the suggestions in the thread bring you to.

You will also need the "angle-addition" formula for sine at some point.

9. Aug 21, 2011

### Ry122

Re: parllax

Can you please re-arrange your equation in post 6 so that h is the subject and tell me how it's any different from mine?

10. Aug 21, 2011

### dynamicsolo

Re: parllax

$h = \frac{c}{ (\frac{cos\alpha}{sin\alpha} + \frac{cos\beta}{sin\beta} )}$

c has to be divided by the sum of the two ratios.

11. Aug 21, 2011

### Ry122

Re: parllax

Thanks. Is the angle-addition formula you're referring to the trigonometric identity shown in my first post? I'm not sure how I would apply that here.

12. Aug 21, 2011

### dynamicsolo

Re: parllax

Add the two ratios together and you'll see where it needs to be applied.