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