Trig identity question

  • Thread starter Ed Aboud
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  • #1
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Homework Statement



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

Trying to find [itex] \theta [/itex]

[tex] \frac{1}{2} (sin2 \theta) tan^2 \theta -tan \theta + \frac{1}{2} sin2 \theta = 0[/tex]

Homework Equations





The Attempt at a Solution


I tryed spliting [itex] tan \theta [/itex] into [itex] \frac{sin \theta}{cos \theta} [/itex] 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!!
 

Answers and Replies

  • #2
gabbagabbahey
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The identities [itex]\sin(2\theta)=2\sin(\theta)\cos(\theta)[/itex] and [itex]\sec^2(\theta)=\tan^2(\theta)+1[/itex] should both be useful.

However, you won't be able to solve your equation for [itex]\theta[/itex] since it turns out your equation is true for all [itex]\theta[/itex]!:eek:

You must have made an error earlier in the problem.
 
  • #3
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hmm I got sin x =0 so x=0.
 
  • #4
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[tex] sin \theta cos \theta (tan^2 \theta) - tan \theta + sin \theta cos \theta = 0 [/tex]

[tex] sin \theta cos \theta (sec^2 \theta) - \frac{sin \theta}{cos \theta} = 0 [/tex]

[tex] cos^2 \theta (sec^2 \theta) = 1 [/tex]

[tex] 1 =1 [/tex]

?
 
  • #5
gabbagabbahey
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hmm I got sin x =0 so x=0.

Then you should double check your algebra :wink:
 
  • #6
gabbagabbahey
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[tex] sin \theta cos \theta (tan^2 \theta) - tan \theta + sin \theta cos \theta = 0 [/tex]

[tex] sin \theta cos \theta (sec^2 \theta) - \frac{sin \theta}{cos \theta} = 0 [/tex]

[tex] cos^2 \theta (sec^2 \theta) = 1 [/tex]

[tex] 1 =1 [/tex]

?

Like I said; your equation is true for all [itex]\theta[/itex], so you must have made an error earlier in the problem.
 
  • #7
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[tex] \frac{1}{2} (sin2 \theta) tan^2 \theta -tan \theta + \frac{1}{2} sin2 \theta = 0[/tex]

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

[tex] sin(\theta)cos(\theta) \frac{sin^{2}(\theta)}{cos^{2}(\theta)} - \frac{sin(\theta)}{cos(\theta)} + sin(\theta)cos(\theta) = 0 [/tex]

Then I found the common denominator and got:

[tex] \frac{sin^{3}(\theta) - sin(\theta) + sin(\theta)cos^{2}(\theta)}{cos(\theta)} = 0 [/tex]

So now we have to consider 2 cases

1) [tex] cos(\theta) = 0 [/tex]

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

2) [tex] cos(\theta) \neq 0 \Rightarrow sin(\theta)(sin^{2}(\theta) - 1 + cos^{2}(\theta)) = 0 [/tex]

The 2nd case gives you 0 = 0 so it seems like your expression holds true for any x.
 
  • #8
446
1

Homework Statement



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

Trying to find [itex] \theta [/itex]

[tex] \frac{1}{2} (sin2 \theta) tan^2 \theta -tan \theta + \frac{1}{2} sin2 \theta = 0[/tex]

Homework Equations





The Attempt at a Solution


I tryed spliting [itex] tan \theta [/itex] into [itex] \frac{sin \theta}{cos \theta} [/itex] 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:
  • #9
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Then you should double check your algebra :wink:

lol but if you put zero degree in you do get 0. I have probability done something wrong.
 
  • #10
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Zero degree? Are you saying if you evaluate it at theta = 0 you get 0 = 0? Did you read the post I made?
 
  • #11
gabbagabbahey
Homework Helper
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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?:wink:
 
  • #12
344
1
First thing I did was find rewrite everything in terms of sin and cos:

[tex] sin(\theta)cos(\theta) \frac{sin^{2}(\theta)}{cos^{2}(\theta)} - \frac{sin(\theta)}{cos(\theta)} + sin(\theta)cos(\theta) = 0 [/tex]

Then I found the common denominator and got:

[tex] \frac{sin^{3}(\theta) - sin(\theta) + sin(\theta)cos^{2}(\theta)}{cos(\theta)} = 0 [/tex]

So now we have to consider 2 cases

1) [tex] cos(\theta) = 0 [/tex]

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

2) [tex] cos(\theta) \neq 0 \Rightarrow sin(\theta)(sin^{2}(\theta) - 1 + cos^{2}(\theta)) = 0 [/tex]

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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