Solving for Theta: Help With a Ballistics Equation

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    Ballistics Theta
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Discussion Overview

The discussion revolves around solving for the angle theta in a ballistics equation involving trigonometric functions. Participants explore the relationship between the trigonometric identity and the given equation, focusing on how to isolate theta.

Discussion Character

  • Homework-related, Mathematical reasoning

Main Points Raised

  • One participant presents the equation cos(theta)*sin(theta) = gd/2v^2 and seeks assistance in solving for theta.
  • Another participant suggests using the identity cos(t)sin(t) = sin(2t)/2 to facilitate solving for theta.
  • A later reply reiterates the trigonometric identity for clarity, emphasizing the transformation of the original equation.

Areas of Agreement / Disagreement

Participants appear to agree on the use of the trigonometric identity but do not reach a consensus on the method to isolate theta from the equation.

Contextual Notes

The discussion does not address potential limitations in the assumptions made about the variables involved or the specific conditions under which the trigonometric identities apply.

VantagePoint72
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Hey everyone,
I'm trying to solve for theta in a derivative of a ballistics equation and I'm afraid I'm stuck. My trigonometry is a little rusty, can someone help me out? I have:

cos(theta)*sin(theta)=gd/2v^2

With the information I'm given I can solve the the right side of the equation ending up with:

cos(theta)*sin(theta)=some number

How do I get theta by knowing what cos(theta)*sin(theta) equals?

Thanks!
 
Physics news on Phys.org
cos(t)sin(t) = 2sin(t)cos(t)/2 = sin(2t)/2. Now you can use arcsine...
 
Just clarifying what Muzza wrote above...so there's no room for misunderstanding :

[tex]cos(t)sin(t) = \frac {2sin(t)cos(t)}{2} = \frac {sin(2t)}{2}.[/tex]
 
OK, thanks for the help!
 

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