How do you evaluate #y=sin(x + (pi/2) )#?

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Answer 1 · 2016-08-01T16:18:32

#y=cos x#

#y=sin(x+(pi/2)) ?#

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

#y=sin x*cos (pi/2)+cos x*sin(pi/2)#

#"plug "sin (pi/2)=1" ; "cos(pi/2)=0#

#y=sin x*0 +cos x*1#

#y=0+cos x#

#y=cos x#