How do you prove #sin ((5pi)/4)#?

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Answer 1 · 2016-06-25T15:23:18

#sin((5pi)/4)=-sqrt2/2#

#sin((5pi)/4)=sin(pi+pi/4)#

#sin(a+b)=sin a*cos b+cos a*sin b#

#sin(pi+pi/4)=sin pi*cos (pi/4)+cos pi*sin(pi/4)#

#sin pi=0" ; "cos pi=-1#

#sin(pi/4)=sqrt 2/2" ; "cos (pi/4)=sqrt2/2#

#sin((5pi)/4)=0*sqrt2/2-1*sqrt2/2#

#sin((5pi)/4)=-sqrt2/2#