Use expansion of #sin(A+B)# to prove #sin(x+pi/2)=cosx# ?

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Answer 1 · 2017-05-25T05:55:38

See the proof below

We need

#sin(A+B)=sinAcosB+sinBcosA#

#cos(1/2pi)=0#

#sin(1/2pi)=1#

Therefore,

#sin(x+1/2pi)=sinxcos(1/2pi)+cosxsin(1/2pi)#

#=sinx*0+cosx*1#

#=cosx#

#QED#