How do you evaluate sine, cosine, tangent of #(10pi)/3# without using a calculator?

1 Answer
Andrea S.
Dec 28, 2016

#sin((10pi)/3)= -sqrt(3)/2#

#cos((10pi)/3)= -1/2#

#tan((10pi)/3)= sqrt(3)#

Explanation:

You consider that: #(10pi)/3 = 2pi+pi+pi/3#, so that:

#sin((10pi)/3)= sin(2pi+pi+pi/3) =sin(pi+pi/3) = sinpicos(pi/3)+cospisin(pi/3) = -sin(pi/3)= -sqrt(3)/2#

#cos((10pi)/3)= cos(2pi+pi+pi/3) =cos(pi+pi/3) = cospicos(pi/3)-sinpisin(pi/3) = -cos(pi/3)= -1/2#

#tan((10pi)/3)= sin((10pi)/3)/(cos((10pi)/3))=(-sqrt(3)/2)/(-1/2) = sqrt(3)#

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