Trig identity question

Ed Aboud

1. Homework Statement

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

Trying to find $\theta$

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

2. Homework Equations

3. 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!!

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gabbagabbahey

Homework Helper
Gold Member
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

Homework Helper
Gold Member
hmm I got sin x =0 so x=0.
Then you should double check your algebra gabbagabbahey

Homework Helper
Gold Member
$$sin \theta cos \theta (tan^2 \theta) - tan \theta + sin \theta cos \theta = 0$$

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

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

$$1 =1$$

?
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

1. Homework Statement

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

Trying to find $\theta$

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

2. Homework Equations

3. 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

Last edited:

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

Homework Helper
Gold Member
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!

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