Phasor Addition Simplify 5cos(wot+90°)+5cos(wot-30°)+5cos(wot-120°): Solve

[noname1]

simplfy 5cos(wot+90°)+5cos(wot-30°)+5cos(wot-120°) to form Acos(cos+Ø)

i did it this way
5cos(wt) = 5e^j0 = 5+J0
5cos(wot-30°) = 5e^-j30° = 5*(√3/2)+J5/2
5cos(wot-120°) = 5e^-j120° = -5/2*(√3/2)+J5*(√3/2)

i have only done partially the problem just missing to add them up however when i look up the solutions this portion where i convert it to Cartesian it doesn't match the solutions but i quite positive i am right, could anyone confirm?

 

[The Electrician]

simplfy 5cos(wot+90°)+5cos(wot-30°)+5cos(wot-120°) to form Acos(cos+Ø)

i did it this way
5cos(wt) = 5e^j0 = 5+J0
5cos(wot-30°) = 5e^-j30° = 5*(√3/2)+J5/2
5cos(wot-120°) = 5e^-j120° = -5/2*(√3/2)+J5*(√3/2)

i have only done partially the problem just missing to add them up however when i look up the solutions this portion where i convert it to Cartesian it doesn't match the solutions but i quite positive i am right, could anyone confirm?

I think the part in red needs to be changed; it should be:

-5/2 - j5*(√3/2)

 

[noname1]

I think the part in red needs to be changed; it should be:

-5/2 - j5*(√3/2)

yes that is correct it was a typo but the rest looks correct right?

than adding these up 5+5*√(3)/2-(5/2) = 6.83
j0+j5/2-j5√(3)/2 = -j1.83

than magnitude would be √6.83²+1.83² = 7.07
and tan^-1 = (-1.83/6.83) = -14.999° = -.08pi

 

[noname1]

nevermind i found the issue but thanks

 

[rezeew]

Yes, your approach and calculations seem correct. However, it would be helpful to see the full solution and compare it to the given solution to confirm if there are any errors or differences. Additionally, it is always a good idea to double check your calculations and equations to ensure accuracy.