Question

How do you solve `sin(theta + pi/6) - cos (theta + pi/3) = sqrt3 * sin theta`?

Answer 1

Prove trig expression

Explanation:

Apply the 2 trig identities:
sin (a + b) = sin a.cos b + sin b.cos a
cos (a + b) = cos a.cos b - sin a.sin b
`sin (t + pi/6) = sin t.cos (pi/6) + sin (pi/6).cos t =`
`= (sqrt3/2).sin t + (cos t)/2` (1)
`cos (t + pi/3) = cos (t).cos (pi/3) - sin (pi/3).sin t =`
`= cos (t/2) - (sqrt3/2).sin t.` (2)
Subtract (2) from (1), we get:
`2(sqrt3/2).sin t = sqrt3.sin t`