Understanding the Relationship between y(t) and Ccos[ω(t)-σ]

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Discussion Overview

The discussion revolves around the mathematical relationship between the expressions y(t) = A cos[ω(t)] + B sin[ω(t)] and C cos[ω(t) - σ], exploring the conditions under which they are equivalent. The focus is on the derivation and transformation of these trigonometric expressions.

Discussion Character

  • Mathematical reasoning

Main Points Raised

  • One participant questions why y(t) can equal C cos[ω(t) - σ] and provides the definitions of C and σ.
  • Another participant suggests using the cosine angle subtraction formula to expand C cos(ω(t) - σ) to explore the equivalence.
  • A participant reports their expansion of C cos(ω(t) - σ) and seeks confirmation on whether they are missing a step.
  • Another participant proposes defining A and B in terms of C and σ to further analyze the relationship.
  • A participant concludes that their manipulation leads back to the original equation, affirming the equivalence.
  • It is noted that A² + B² = C² and tan(σ) = B/A, reinforcing the relationships among the variables.

Areas of Agreement / Disagreement

Participants appear to agree on the mathematical transformations and relationships presented, but the discussion does not indicate any explicit disagreements or unresolved issues.

Contextual Notes

The discussion relies on the definitions of A, B, C, and σ, and assumes familiarity with trigonometric identities and manipulations. No limitations or unresolved steps are explicitly mentioned.

Who May Find This Useful

This discussion may be useful for individuals interested in trigonometric identities, mathematical transformations, or those studying relationships between different forms of wave equations.

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why can y(t) = A cos[omega(t)]+bsin[omega(t)]
equal Ccos[omega(t)-sigma], where C=(A^2 + B^2)^(1/2) and
tan(sigma)=B/A?
 
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Well

[tex]\cos(a-b) = \cos a \cos b + \sin a \sin b[/tex]

Try expanding out [itex]C\cos(\omega t - \sigma)[/itex] using that formula, and see what you get.
 
i got C[cos(wt)cos(sigma)+sin(wt)sin(sigma)]...
am i missing a step?
@@a
 
Now define A=Ccos(sigma) and B=Csin(sigma).

What do you get if you eliminate sigma from these two equations?
 
C[cos(wt)A/C + sin(wt)B/C]= the orignal equation!
thanks!
 
Note also that [itex]A^2 + B^2 = C^2[/itex] and tan(sigma) = B/A, with the given definitions.
 
cool! thanks again! :)
 

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