How do you evaluate #sin ((11pi)/2)#?

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Answer 1 · 2016-02-28T07:01:48

#-#1

sin (2n + 1)#pi#/2 = (#-#1) ^ n, for any integer n.,
Here, n = 5 and (#-#1)^5 = #-#1.

Answer 2 · 2016-02-28T07:02:46

#sin(11pi/2)=-1#

#sin(11pi/2)=sin(6pi-(pi)/2)=sin(-(pi)/2)#

Note #6pi# is a multiple of #2pi#

As #sin(-pi/2)=-1#

#sin(11pi/2)=-1#