How do you prove #cos(x + pi/6) + sin(x + pi/3) = (sqrt 3)cos x#?

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Answer 1 · 2016-03-19T09:05:03

We use some basics like
#Cos(90-A) = sinA#

#sin A + sin B = 2 sin (A + B)/2 cos(A - B)/2#

let me rewrite the given LHS;

#sin(x + (pi)/3) + cos(x + x/6)#

#sin(x + (pi)/3) + sin(pi/2-(x + pi/6))#

#sin(x + (pi)/3) + sin(pi/2-x - pi/6)#

#sin(x + (pi)/3) + sin(pi/3-x)#

Now recall

#sin A + sin B = 2 sin (A + B)/2 cos(A - B)/2#

we get

#=2 sin(( (x + (pi)/3) + (pi/3-x))/2) cos(( (x + (pi)/3) - (pi/3-x))/2)#

#=2 sin (pi/3) cos x#

#=2 * sqrt3/2 * cos x#

#=cancel2 * sqrt3/cancel2 * cos x#

#= sqrt3 cos x = RHS#

QED