Evaluate the trigonometric sum sin(pi/3)+sin(2pi/3)+sin(3pi/3)+sin(4pi/3) .
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Since this trigonometric sum involves the angles that are multiples of pi/3 (equivalent to 60 degrees)—a "special" angle with known sine, cosine, and tangent values—each of the terms can be evaluated separately.
- pi/3 is an acute angle located in the first quadrant, and it has a known sine value: sin(pi/3) = sqrt(3)/2
- 2pi/3 is located in the second quadrant, so its sine value is positive and equal to that of pi/3: sin(2pi/3) = sqrt(3)/2
- 3pi/3 = pi. This is a quadrantal angle, located on the x-axis, so its sine is zero: sin(3pi/3) = 0
- 4pi/3 is located in the third quadrant, so its sine value is negative and equal to the opposite of that of pi/3: sin(4pi/3) = -sqrt(3)/2
Adding it all together, we get sqrt(3)/2 + sqrt(3)/2 + 0 + (-sqrt(3)/2) = sqrt(3)/2.
The value of the trigonometric sum is sqrt(3)/2.
Please see the linked website for more information about the unit circle.
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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.
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