Not sure why cos(phi) = sin(phi) = 1

Thread Starter

raddian

Joined Jun 20, 2015
6
Screenshot - 06252015 - 07:38:57 AM.png
[Taken from Basic Engineering Circuit Analysis, 10th Ed. Irwin.]

\(
\begin{align}
i(t) &= A \cos(\omega t+ \phi)
i(t) &=A \cos(\phi)\cos(\omega t) - A \sin(\phi) \sin(\omega t)
\end{align}
i(t) =A \cos(\omega t) + A \sin(\omega t)
\)
And they decided to just make cos(phi) = 1 = sin(phi), Why did this happen?

(here is the rest of the example, which I dont think provides explanation)
Screenshot - 06252015 - 07:46:50 AM.png
 
Last edited:

Papabravo

Joined Feb 24, 2006
21,094
The only value of φ for which they are equal is π/4 or 45°. They can never be simultaneously equal to 1 -- EVER!
It is true that φ is a constant, and it is true that A is a constant, and it is also true that
\(A_1 and A_2 \)
are constants, but they don't have to be equal.
 
Last edited:

MrAl

Joined Jun 17, 2014
11,342
Hi,

Yes that is the perfect reason for the misunderstanding Wbahn.

This is a trigonometric identity that is often used in electronics but written sometimes in a slightly more apparent form is:
A*cos(wt)+B*sin(wt)=C*cos(wt+ph)

or alternately:
C*cos(wt+ph)=A*cos(wt)+B*sin(wt)

You're goal is either to find A and B, or to find C and ph, depending on which way you are going, in order to either get rid of the phase shift or to include a phase shift in the final form.
Writing it this way makes it a little more apparent, but then we just have C equal to A in the previous form, and A1 and A2 equal to A and B in the second form. So it's the same thing just written using A, B, and C rather than A1, A2, and A.

So the basic idea is to either get rid of the phase shift or to convert to a form with a phase shift. The phase shift can also be calculated, given the two term form, as well as the constant C.
 
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