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How do you evaluate #sin ((-5pi)/6)#?

1 Answer

Shwetank Mauria

Oct 28, 2016

#sin(-(5pi)/6)=-1/2#

Explanation:

As #sin(-theta)=-sintheta# and #sin(pi-theta)=sintheta#

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

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

= #-sin(pi/6)#

= #-1/2#

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