Evaluate the trigonometric sum sin(pi/3)+sin(2pi/3)+sin(3pi/3)+sin(4pi/3) .

We'll group the 1st and the last terms and the middle terms
together.

[sin(pi/3)+sin(4pi/3)]+[sin(2pi/3)+sin(3pi/3)]

Since
the function inside brackets are matching, we'll transform them into
products.

S=2sin[(pi/3+4pi/3)/2]*cos[(pi/3-4pi/3)/2]+2sin[(2pi/3+3pi/3)/2]*cos[(2pi/3-3pi/3)/2]

S=2sin(5pi/6)*cos(pi/2)+2sin(5pi/6)*cos(pi/6)

But
cos(pi/2)=0

S=2sin(5pi/6)*cos(pi/6)

We'll
write sin(5pi/6) = sin(6pi/6 -
pi/6)

sin(5pi/6)=sin(pi-pi/6)

sin(5pi/6)=sin(pi/6)

S=2sin(pi/6)cos(pi/6)

We'll
recognize the double angle
identity:

S=sin2*(pi/6)=sin(pi/3)

S=sqrt3/2

The
value of trigonometric sum is S=sqrt3/2.