I'm not sure if this is considered pre- or post-calculus (I am doing it for a calculus course, but I doubt it involves actual calculus) but I'll go ahead and post here. 1. The problem statement, all variables and given/known data I am required to re-write the function sin x + 2 cos (x - pi/6) in the form r sin (x + theta). 2. Relevant equations cos (x) = sin (x + pi/2) perhaps equations for sin (a + b) and cos (a + b)? 3. The attempt at a solution So I tried to get everything to be sine: sin x + 2 cos (x - pi/6) = sin x + 2 sin (x - pi/6 + pi/2) = sin x + 2 sin (x + pi/3) I now tried to expand the second term using the equation sin(a+b) = sin(a)cos(b) + cos(a)sin(b): = sin x + 2 [ sin (x) cos(pi/3) + cos (x) sin (pi/3)] = sin x + 2 [sin(x)(1/2) + cos(x)(sqrt(3)/2)] = sin x + sin x + sqrt(3) cos x = 2 sin x + sqrt(3) cos x But I don't know where to go from here. Trying to convert this new cosine to sine using cos (x) = sin (x + pi/2) just brings you back to the same thing after simplifying. So where do I go from here to get it to be in the form r sin (x + theta)?