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

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We have to find the result of sin(pi/3) + sin(2pi/3) + sin(3pi/3) + sin(4pi/3)

We use sin (pi/3) = (sqrt 3)/2

sin (2*pi/3) = sin (pi - pi/3) = sin (pi/3) = (sqrt 3)/2

sin (3*pi/3) = sin (pi) = 0

sin (4*pi/3) = sin (pi/3 + pi) = -sin (pi/3) = -(sqrt 3)/2

Adding (sqrt 3)/2 + (sqrt 3)/2 + 0 - (sqrt 3)/2

=> (sqrt 3)/2

**The required sum of sin(pi/3) + sin(2pi/3) + sin(3pi/3) + sin(4pi/3) = (sqrt 3)/2**

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.**