Transform this function into phasor form

[Moneer81]

Tansform the following function of time into phasor form:

3 cos 600t - 5 sin (600t + 110)

so I need to change the second term to a cosine term, any hints? or can I use a trig indentity that would make things easier?

 

[malawi_glenn]

There are a lot of formulas that can help you. I my self often uses the Euler identities.

 

[Architeuthis Dux]

Yes, you can use a trigonometric identity to make things easier. Specifically, you can use the identity sin(x + y) = sin(x)cos(y) + cos(x)sin(y). This means that you can rewrite the second term as:

-5 sin (600t + 110) = -5(sin(600t)cos(110) + cos(600t)sin(110))

Then, using the phasor form of sine and cosine functions, which are defined as:

cos(ωt) = Re{e^(jωt)} = (1/2)(e^(jωt) + e^(-jωt))

sin(ωt) = Im{e^(jωt)} = (1/2j)(e^(jωt) - e^(-jωt))

Where ω is the angular frequency and j is the imaginary unit. Substituting these into the equation, we get:

-5(sin(600t)cos(110) + cos(600t)sin(110)) = -5((1/2)(e^(j600t) + e^(-j600t))cos(110) + (1/2j)(e^(j600t) - e^(-j600t))sin(110))

= -5((1/2)(e^(j600t) + e^(-j600t))cos(110) + (1/2j)(e^(j600t)sin(110) - e^(-j600t)sin(110)))

= -5((1/2)(e^(j600t) + e^(-j600t))cos(110) + (1/2j)(e^(j600t + j110) - e^(j600t - j110)))

= -5((1/2)(e^(j600t) + e^(-j600t))cos(110) + (1/2j)(e^(j600t + j110) - e^(j600t - j110)))

= -5((1/2)(e^(j600t) + e^(-j600t))cos(110) + (1/2j)(e^(j600t + j110) - e^(j600t - j110)))

= -5((1/2)(e^(j600t) + e^(-j