How do you solve sin(theta + pi/6) - cos (theta + pi/3) = sqrt3 * sin thetasin(θ+π6)cos(θ+π3)=3sinθ?

1 Answer
Apr 4, 2016

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 =sin(t+π6)=sint.cos(π6)+sin(π6).cost=
= (sqrt3/2).sin t + (cos t)/2=(32).sint+cost2 (1)
cos (t + pi/3) = cos (t).cos (pi/3) - sin (pi/3).sin t =cos(t+π3)=cos(t).cos(π3)sin(π3).sint=
= cos (t/2) - (sqrt3/2).sin t.=cos(t2)(32).sint. (2)
Subtract (2) from (1), we get:
2(sqrt3/2).sin t = sqrt3.sin t2(32).sint=3.sint