Conversion to phasor from cos and sine

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The discussion focuses on converting the expression 100Sin(ωt+30) + 20cos(ωt) into phasor form, resulting in 78.7<38.9 degrees. The key relationship used for this conversion is √2Esin(ωt-∅)=Ee^(-j∅). The initial approach involved using trigonometric identities to express both terms as sine functions, but the user encountered difficulties before successfully solving the problem.

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vvl92
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I am given the problem:
100Sin(ωt+30) +20cos(ωt)

and the solution is 78.7<38.9degrees

How do I convert to phasor form?

I know for this calculation I need the following relationship:
√2Esin(ωt-∅)=Ee^(-j∅)

At first I tried making both sines and then splitting up 100Sin(ωt+30) and 20Sin(ωt+90)using the trigonometric identities but I get stuck at that. Please help!
 
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Solved it :D
Please let me know if anyone would like me to go over it.
 

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