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!