Express 3sin(2x) + 5cos(2x) in the form Rsin(2x+a), R>0 0<a<pi/2

Start by expanding out Rsin(2x+a) using the addition formula for sin, sin(A+B) = sin(A)cos(B)+cos(A)sin(B). Substituting 2x = A and a= b, we get that Rsin(2x+a) = R(sin(2x)cos(a) + cos(2x)sin(a)) = Rcos(a)sin(2x) + Rsin(a)cos(2x). Equating this to the original expression 3sin(2x) + 5cos(2x), then gives that Rcos(a) = 3 and Rsin(a) = 5. Squaring these equations then adding them gives 9 + 25 = 34 = R²(sin²(a) + cos²(a)) = R² by using the trigonometric identity sin²(a) + cos²(a) = 1. It then follows that R = sqrt(34). This value of R can then be substituted into Rcos(a) = 3 to give that cos(a) = 3/sqrt(34) and then a = arccos(3/sqrt(34)) = 1.03 to 3 sig. fig. This can be checked by taking any value of 2x , say 2, then plugging the numbers that you get into a calculator, you should get 3sin(2) + 5cos(2) - sqrt(34)sin(2+1.03) = 0