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

Question

3389 views around the world

You can reuse this answer
Creative Commons License

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)#