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

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The discussion explores the relationship between the equations y(t) = A cos[ω(t)] + B sin[ω(t)] and C cos[ω(t) - σ], establishing that C = (A^2 + B^2)^(1/2) and tan(σ) = B/A. Participants clarify the expansion of C cos(ω(t) - σ) using the cosine angle subtraction formula, leading to a confirmation of the original equation. By defining A and B in terms of C and σ, they demonstrate that the two forms are equivalent. The relationship A^2 + B^2 = C^2 is reiterated, reinforcing the mathematical connection. This discussion effectively illustrates the trigonometric identities and their application in transforming sinusoidal functions.
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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

\cos(a-b) = \cos a \cos b + \sin a \sin b

Try expanding out C\cos(\omega t - \sigma) 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 A^2 + B^2 = C^2 and tan(sigma) = B/A, with the given definitions.
 
cool! thanks again! :)
 

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