# Electrical: Reduce the expression to the form V_mcos(wt+theta)

1. Apr 23, 2013

### Northbysouth

1. The problem statement, all variables and given/known data

Reduce the Expression:

15sin(wt-45°) + 5cos(wt-30°) + 10cos(wt-120°)

to the form Vmcos(wt+θ)

2. Relevant equations

3. The attempt at a solution

My theta value at the end isn't coming out right.

My first step was to put make sure each term was in terms of cosine, hence:

15cos(wt-45°-90°) + 5cos(wt-30°) + 10cos(wt-120°)

Which in phasor form is:

15∠-135 + 5∠-30 + 10∠-120

Then I used cos and sin to find the real values and the j values

15cos(-135) + j15sin(-135) + 5cos(-30) + j5sin(-30) + 10cos(-120) + j10sin(-120)

= -11.28 -j21.77

Which in a phasor format is:

24.52∠62.6

24.53 = (11.282+ 21.772)1/2

62.6 = arctan(21.77/11.28)

But apparently the solution is:

24.52∠-117.39°

What am I missing?

Any help would be appreciated. Thank you

2. Apr 23, 2013

### Staff: Mentor

Note the signs of the complex components of your phasor sum. In what quadrant will the phasor lie?

3. Apr 23, 2013

### Northbysouth

Are you saying that because I have -11.28 - j21.77 as negatives that the phasor will be in the third quadrant and consequently the angle made with the postive x-axis will be 180-62.6?

4. Apr 23, 2013

### Staff: Mentor

The phasor will be in the third quadrant, yes. You can either add 180 or subtract 180, but generally the convention is to choose the result with a magnitude less than or equal to 180. Alternatively, employ the atan2(y,x) function rather than the atan function (if your calculator has it); it automatically sorts out the quadrant from the given y,x data.