# Math Help - Solving two trig equations

1. ## Solving two trig equations

So I'm trying to find the horizontal and vertical tangents to a polar curve, and what it comes down to is solving the following equations:

4cos(4x)sin(x) + sin(4x)cos(x) = 0
and
4cos(4x)cos(x) - sin(4x)sin(x) = 0

I'm fairly lost; I keep messing around with the equations, but it doesn't seem to get any simpler. Anything I do to it ends up being equally complex. Any suggestions?

2. ## Re: Solving two trig equations

I forgot to mention this is over the interval [0,2pi). For 4cos(4x)sin(x) + sin(4x)cos(x) I can easily see that at 0 and pi the sin portions will produce solutions, but I can't think of a way to find where else
4cos(4x)sin(x) = -sin(4x)cos(x)
especially since the cos(4x) and cos(x) will never both equal 0 for the same angle. I have tried reducing cos(4x) and sin(4x) down using half angle formulas, but the end result just seems more messy. I was able to change both equations into forms of tan, with the first equation being:
-tan(4x) = 4tan(x)
This just feels like a step in the wrong direction though.

3. ## Re: Solving two trig equations

Each of these functions have plenty of zeros on [0,2pi) but they don't seem to have any in common. Do you need to solve these equations simultaneously? If so there is no solution.

You might be able to take tan(4x) and express it in terms of sin(x) and cos(x), it looks pretty nasty though. Are you sure the factor of 4 at the front of each is correct? If that wasn't there this would just be expressions for sin(5x) and cos(5x). Perhaps post some of your work that led up to these two equations.

4. ## Re: Solving two trig equations

Oh sorry no I am not trying to solve them simultaneously. Here is what I did leading up to it. This is a calculus problem, but it ends up just being a trig problem in the end. I had put x but it is really theta so I'll switch back to that to avoid confusion.
I started trying to find the vertical and horizontal tangents of the polar curve:

$r = \sin(4\theta)$
$x = r\cos(\theta)$
$y = r\sin(\theta)$

$\frac{dy}{dx} = \frac{\frac{d}{d\theta}(r\sin(\theta))}{\frac{d}{d \theta}(r\cos(\theta))} = \frac{\frac{d}{d\theta}(\sin(4\theta)\sin(\theta)) }{\frac{d}{d\theta}(\sin(4\theta)\cos(\theta))} = \frac{4\cos(4\theta)\sin(\theta) + \sin(4\theta)\cos(\theta)}{4\cos(4\theta)\cos(\the ta) - \sin(4\theta)\sin(\theta)}$

so you see I am left with two different trig equations to solve,
$4\cos(4\theta)\sin(\theta) + \sin(4\theta)\cos(\theta) = 0$
and
$4\cos(4\theta)\cos(\theta) - \sin(4\theta)\sin(\theta) = 0$

I guess I should probably repost this in the calculus forum now that I think about it.

5. ## Re: Solving two trig equations

I did a pretty picture
The red on is the the first one
The blue on is the second one.
I only just discovered gyazo today so I cannot resist the temptation of seeing it work.

f7e9c0ba7cab91ca74e58c4baeff48d6.gif

6. ## Re: Solving two trig equations

I can show you what I was able to do

$4 \sin (q) \cos (4 q)+\sin (4 q) \cos (q)$

=

$\frac{5 \sin ^5(q)}{2}+\frac{3 \sin ^3(q)}{2}+\frac{25}{2} \sin (q) \cos^4(q)-25 \sin ^3(q) \cos ^2(q)-\frac{9}{2} \sin (q) \cos ^2(q)$

Now substitute cos2(q) terms with (1 - sin2(q))

$\frac{5 \sin ^5(q)}{2}+\frac{3 \sin ^3(q)}{2}-\frac{9}{2} \left(1-\sin^2(q)\right) \sin (q)-25 \left(1-\sin ^2(q)\right) \sin^3(q)+\frac{25}{2} \sin (q) \cos ^4(q)$

and cos4(q) terms with (1 - sin2(x))2

$\frac{5 \sin ^5(q)}{2}+\frac{3 \sin ^3(q)}{2}+\frac{25}{2} \left(1-\sin^2(q)\right)^2 \sin (q)-\frac{9}{2} \left(1-\sin ^2(q)\right) \sin(q)-25 \left(1-\sin ^2(q)\right) \sin ^3(q)$

and simplify it all (let mathematica simplify it )

$\frac{1}{2} (5 \sin (5 q)-3 \sin (3 q))$

setting above to 0 you get

$5 \sin (5 q)=3 \sin (3 q)$

It's not particularly solvable but it's a remarkable form.

If you do the same thing with the denominator you obtain

$\frac{1}{2} (3 \cos (3 q)+5 \cos (5 q))$

and setting to 0 you get

$3 \cos (3 q)=-5 \cos (5 q)$

ok you can play with it from here.

7. ## Re: Solving two trig equations

Wow it's ridiculous that it can be simplified down so much, thanks a ton.