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