Why is this equation equal to another equation?

  • Context: Undergrad 
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Discussion Overview

The discussion centers on the relationship between two equations involving trigonometric functions, specifically how the expression for distance in projectile motion can be transformed using trigonometric identities. The scope includes mathematical reasoning and conceptual clarification of trigonometric identities.

Discussion Character

  • Mathematical reasoning

Main Points Raised

  • One participant presents the equation x = (2vsin(theta)cos(theta))/g and asks how it is equivalent to x = (vsin2(theta))/g, expressing confusion about the disappearance of cos(theta) and the emergence of sin2(theta).
  • Another participant cites the trigonometric identity 2sin(theta)cos(theta) = sin(2theta) as the basis for the equivalence between the two equations.
  • A later reply reiterates the same trigonometric identity and expands on it by mentioning additional identities related to sine and cosine.

Areas of Agreement / Disagreement

Participants generally agree on the validity of the trigonometric identity used to relate the two equations, but the initial participant's confusion about the transformation remains unaddressed, indicating some unresolved aspects of the discussion.

Contextual Notes

The discussion does not clarify the implications of using the identity in the context of the original equations, nor does it address any assumptions about the variables involved, such as the definition of null velocity.

Who May Find This Useful

Individuals interested in trigonometric identities, projectile motion equations, or mathematical transformations may find this discussion relevant.

mapa
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x = (2vsin(theta)cos(theta))/g = (vsin2(theta))/g

v = null velocity

x = distance

g = gravity

How is the first equation equal to the second one?
How the sin2(theta) come to be and where did the cos(theta) go?
 
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Trigonometric identity:

[tex]2 \sin (\theta) \cos (\theta) = \sin (2 \theta)[/tex]
 
thank you
 
Nabeshin said:
Trigonometric identity:

[tex]2 \sin (\theta) \cos (\theta) = \sin (2 \theta)[/tex]

Or more generally, sin(A+B)=sin(A)cos(B)+cos(A)sin(B).
Also handy to know: cos(A+B)=cos(A)cos(B)-sin(A)sin(B).
 

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