How do you prove $sin[x+(pi/6)] + sin[x-(pi/6)] = (sqrt3)/2$ ?

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Answer 1 · 2015-11-12T08:42:10

$sin(x + pi/6) + sin(x - pi/6) = sqrt(3)sin(x)$

Using the trigonometric identity
$sin(alpha + beta) = sin(alpha)cos(beta) + cos(alpha)sin(beta)$

$sin(x + pi/6) = sin(x)cos(pi/6) + cos(x)sin(pi/6)$
and
$sin(x - pi/6) = sin(x)cos(-pi/6) + cos(x)sin(-pi/6)$

As the sine function is odd ( $sin(-x) = -sin(x)$ ) and the cosine function is even ( $cos(-x) = cos(x)$ ), we get

$sin(x - pi/6) = sin(x)cos(pi/6) - cos(x)sin(pi/6)$

Thus, adding gives us

$sin(x + pi/6) + sin(x - pi/6) = 2sin(x)cos(pi/6)$

As $cos(pi/6) = sqrt(3)/2$ we get the result

$sin(x + pi/6) + sin(x - pi/6) = sqrt(3)sin(x)$

How do you prove sin[x+(pi/6)] + sin[x-(pi/6)] = (sqrt3)/2sin[x+(pi/6)] + sin[x-(pi/6)] = (sqrt3)/2?