1. ## Eliminating theta

I have the two equations:

1) x + y = qsin(theta)

2) z = rcos(theta)

The question is eliminate theta and solve for q in terms of x, y, z, and r.

I'm not sure what eliminating theta means, but I have:

arcsin((x+y)/q) = theta

and I substituted that into my top equation. I have no clue on what I should do now.

Also,I don't understand how I can solve for q in terms of z and r when q isn't in the equation with z and r.

2. ## Re: Eliminating theta

How about using the fact that $sin^2(\theta)+ cos^2(\theta)= 1$?

3. ## Re: Eliminating theta

Perhaps an easier approach would be to write your equations as:

(1) $\frac{x+y}{q}=\sin\theta$

(2) $\frac{z}{r}=\cos\theta$

Now square both equations and add, and the right side will be an identity which will eliminate $\theta$.

4. ## Re: Eliminating theta

Should I have sin^2(arcsin(x+y)/q) + cos^2(arccos(z/r)) =1?

5. ## Re: Eliminating theta

Nevermind my last post

6. ## Re: Eliminating theta

I have q^2 = ((x+y)^2)/(1-(z/r)^2)

Sorry, I don't know how to make it look neater than that.

7. ## Re: Eliminating theta

Well, you could write the left side as $\frac{(x+ y)^3}{1- \frac{z^2}{r^2}}= \frac{r^2(x+ y)^2}{r^2- z^2}$, multiplying both numerator and denominator by $r^2$. And, of course, since you are asked to find q, you still need to take the square root of both sides.

8. ## Re: Eliminating theta

$q=\sqrt{\frac{(x+y)^2}{1-\frac{z^2}{r^2}}$

Your method gets rid of the fraction in the denominator so:

$q=\sqrt{\frac{r^2(x+ y)^2}{r^2- z^2}}$

Correct?

9. ## Re: Eliminating theta

Originally Posted by amthomasjr

$q=\sqrt{\frac{(x+y)^2}{1-\frac{z^2}{r^2}}$

Your method gets rid of the fraction in the denominator so:

$q=\sqrt{\frac{r^2(x+ y)^2}{r^2- z^2}}$

Correct?
that looks right.. it can be further simplified as $q=\frac{r(x+ y)}{\sqrt{r^2- z^2}}$

10. ## Re: Eliminating theta

Thank you all.