Question 1115565
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V1 + V2 = 3*cos(5t) + 4*sin(5t) = {{{5*((3/5)*cos(5t) + (4/5)*sin(5t))}}}    (1)


          Since {{{(3/5)^2+(4/5)^2}}} = {{{25/25}}} = 1,  there is (= does exist) the angle "a" in QI such that  sin(a) = {{{3/5}}},  cos(a) = {{{4/5}}}.

          With these identities, we can continue  (1) in this way:




V1 + V2 = 3*cos(5t) + 4*sin(5t) = {{{5*((3/5)*cos(5t) + (4/5)*sin(5t))}}} = 5*(cos(5t)*sin(a) + sin(5t)*cos(a))     (2)

         Now apply the sine addition formula, and you can continue (2) in this way  




V1 + V2 = {{{3*cos(5t) + 4*sin(5t)}}} = {{{5*((3/5)*cos(5t) + (4/5)*sin(5t))}}} = 5*(cos(5t)*sin(a) + sin(5t)*cos(a)) = 5*sin(5t+a).



It is just what the problem wants of you with  A = 5  and  {{{alpha}}} = a = {{{arcsin(3/5)}}} = {{{arccos(4/5)}}}.
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Solved.