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

1 Answer

Mar 15, 2017

#y=cos(x)#

Explanation:

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

Hence:
#sin(x+pi/2) = sin(x)cos(pi/2) + cos(x)sin(pi/2)#

#= sinx*0 + cosx*1 =cosx#

#:.# the graph of #y# is the graph of #cosx# as below:

graph{cosx [-6.243, 6.24, -3.12, 3.123]}

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