Trig Identity Question: Finding theta in Projectile Problem

[Ed Aboud]

Homework Statement

Basically I am finishing of a projectile question and I get stuck here:

Trying to find $\theta$

$$\frac{1}{2} (sin2 \theta) tan^2 \theta -tan \theta + \frac{1}{2} sin2 \theta = 0$$

Homework Equations

The Attempt at a Solution

I tryed spliting $tan \theta$ into $\frac{sin \theta}{cos \theta}$ but I don't really get anywhere.
I know it requires using a trig identity but I can't really see a suitable one.
Thanks for the help in advance!

 

[gabbagabbahey]

The identities $\sin(2\theta)=2\sin(\theta)\cos(\theta)$ and $\sec^2(\theta)=\tan^2(\theta)+1$ should both be useful.

However, you won't be able to solve your equation for $\theta$ since it turns out your equation is true for all $\theta$!

You must have made an error earlier in the problem.

 

[glueball8]

hmm I got sin x =0 so x=0.

 

[Ed Aboud]

$$sin \theta cos \theta (tan^2 \theta) - tan \theta + sin \theta cos \theta = 0$$

$$sin \theta cos \theta (sec^2 \theta) - \frac{sin \theta}{cos \theta} = 0$$

$$cos^2 \theta (sec^2 \theta) = 1$$

$$1 =1$$

?

 

[gabbagabbahey]

hmm I got sin x =0 so x=0.

Then you should double check your algebra

 

[gabbagabbahey]

$$sin \theta cos \theta (tan^2 \theta) - tan \theta + sin \theta cos \theta = 0$$

$$sin \theta cos \theta (sec^2 \theta) - \frac{sin \theta}{cos \theta} = 0$$

$$cos^2 \theta (sec^2 \theta) = 1$$

$$1 =1$$

?

Like I said; your equation is true for all $\theta$, so you must have made an error earlier in the problem.

 

[NoMoreExams]

$$\frac{1}{2} (sin2 \theta) tan^2 \theta -tan \theta + \frac{1}{2} sin2 \theta = 0$$

First thing I did was find rewrite everything in terms of sin and cos:

$$sin(\theta)cos(\theta) \frac{sin^{2}(\theta)}{cos^{2}(\theta)} - \frac{sin(\theta)}{cos(\theta)} + sin(\theta)cos(\theta) = 0$$

Then I found the common denominator and got:

$$\frac{sin^{3}(\theta) - sin(\theta) + sin(\theta)cos^{2}(\theta)}{cos(\theta)} = 0$$

So now we have to consider 2 cases

1) $$cos(\theta) = 0$$

That case cannot hold because your original expression has tan in it and therefore it is assumed that $$cos(\theta) \neq 0$$

2) $$cos(\theta) \neq 0 \Rightarrow sin(\theta)(sin^{2}(\theta) - 1 + cos^{2}(\theta)) = 0$$

The 2nd case gives you 0 = 0 so it seems like your expression holds true for any x.

 

[icystrike]

Homework Statement

Basically I am finishing of a projectile question and I get stuck here:

Trying to find $\theta$

$$\frac{1}{2} (sin2 \theta) tan^2 \theta -tan \theta + \frac{1}{2} sin2 \theta = 0$$

Homework Equations

The Attempt at a Solution

I tryed spliting $tan \theta$ into $\frac{sin \theta}{cos \theta}$ but I don't really get anywhere.
I know it requires using a trig identity but I can't really see a suitable one.
Thanks for the help in advance!

try searching you double angle rules for something related to sin2x=2sinxcosx and apply it

 

[glueball8]

Then you should double check your algebra

lol but if you put zero degree in you do get 0. I have probability done something wrong.

 

[NoMoreExams]

Zero degree? Are you saying if you evaluate it at theta = 0 you get 0 = 0? Did you read the post I made?

 

[gabbagabbahey]

lol but if you put zero degree in you do get 0. I have probability done something wrong.

Try plugging in other values too. ...notice anything?

 

[glueball8]

First thing I did was find rewrite everything in terms of sin and cos:

$$sin(\theta)cos(\theta) \frac{sin^{2}(\theta)}{cos^{2}(\theta)} - \frac{sin(\theta)}{cos(\theta)} + sin(\theta)cos(\theta) = 0$$

Then I found the common denominator and got:

$$\frac{sin^{3}(\theta) - sin(\theta) + sin(\theta)cos^{2}(\theta)}{cos(\theta)} = 0$$

So now we have to consider 2 cases

1) $$cos(\theta) = 0$$

That case cannot hold because your original expression has tan in it and therefore it is assumed that $$cos(\theta) \neq 0$$

2) $$cos(\theta) \neq 0 \Rightarrow sin(\theta)(sin^{2}(\theta) - 1 + cos^{2}(\theta)) = 0$$

The 2nd case gives you 0 = 0 so it seems like your expression holds true for any x.

sorry, yep your right!