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

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Answer 1 · 2016-08-01T16:18:32

$y=cos x$

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

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

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

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

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

$y=0+cos x$

$y=cos x$

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