How do you evaluate #sin(( 7pi) / 2)#?

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Answer 1 · 2018-04-15T08:52:16

#sin((7pi)/2)=-1#

We know that,

#color(red)(sin(3pi+theta)=-sintheta#

Here,

#sin((7pi)/2)=sin((6pi+pi)/2)#

#=sin(3pi+pi/2)...toIII^(rd) Quadrant, where, sin is -ve#

#=-sin(pi/2)#

#=-1#

Note:
#(1)theta=pi/2,(5pi)/2,(9pi)/2,(13pi)/2...=>sintheta=1#
#(2)theta=(3pi)/2,(7pi)/2,(11pi)/2,(15pi)/2...=>sintheta=-1#