20/3.14) sin(2t)+(40\(3.14^2)cos(2t) = 7.5cos(2t-22) can someone please explain how?

2019-04-19T02:53:34Z

The point of the excersise is to use the 7.5 cos(2t-122) to solve an eleectrical engineering problem. but my trig is rusty. Any help is appreciated.

alex2019-04-19T03:01:57Z

4.06 cos(2t)+6.37 sin(2t) = √(4.06^2+6.37^2)[cos(2t) cos(a)+sin(2t)sin(a)] , with a = tan^-1(6.37/4.06) = 1 rad
= 7.55 cos(2t - 1)