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.(adsbygoogle = window.adsbygoogle || []).push({});

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)?

**Physics Forums - The Fusion of Science and Community**

# Manipulating sin/cos functions

Have something to add?

- Similar discussions for: Manipulating sin/cos functions

Loading...

**Physics Forums - The Fusion of Science and Community**