sin(theta+pi/6)+cos(theta+pi/3)=cos theta
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#
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