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What are the sine, cosine, and tangent of 5 pi over 3 radians?

User Danit
by
7.4k points

1 Answer

5 votes

sin((5\pi)/(3)) = sine((1)/(2) * (10\pi)/(3))

sin((5\pi)/(3)) = -\sqrt{(1 - cos((10\pi)/(3)))/(2)}

sin((5\pi)/(3)) = -\sqrt{(1 - cos(600))/(2)}

sin((5\pi)/(3)) = -\sqrt{(1 + (1)/(2))/(2)}

sin((5\pi)/(3)) = -\sqrt{(1(1)/(2))/(2)}

sin((5\pi)/(3)) = -\sqrt{(3)/(4)}

sin((5\pi)/(3)) = -(√(3))/(2)


cos((5\pi)/(3)) = cos((1)/(2) * (10\pi)/(3))

cos((5\pi)/(3)) = -\sqrt{(1 + cos((10\pi)/(3)))/(2)}

cos((5\pi)/(3)) = -\sqrt{(1 + cos(600))/(2)}

cos((5\pi)/(3)) = -\sqrt{(1 - (1)/(2))/(2)}

cos((5\pi)/(3)) = -\sqrt{((1)/(2))/(2)

cos((5\pi)/(3)) = -\sqrt{(1)/(4)}

cos((5\pi)/(3)) = -(1)/(2)


tan((5\pi)/(3)) = tan((1)/(2) * (10\pi)/(3))

tan((5\pi)/(3)) = \sqrt{(1 - cos((5\pi)/(3)))/(1 + cos((5\pi)/(3)))}

tan((5\pi)/(3)) = \sqrt{(1 + (1)/(2))/(1 - (1)/(2))}

tan((5\pi)/(3)) = \sqrt{(1(1)/(2))/((1)/(2))}

tan((5\pi)/(3)) = √(3)

tan((5\pi)/(3)) \approx 1.732050808
User Kiswa
by
8.9k points
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