Question 1149958
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Apply the addition formula for sine


    sin(a+b) - sin(a)*cos(b) + cos(a)*sin(b).     (1)


It is one of the basic formula of Trigonometry; it is valid for all angles  "a" and  "b".


Put  a = x  and  b = {{{pi/2}}}.  You will get


    {{{sin(x + pi/2)}}} = {{{sin(x)*cos(pi/2)}}} + {{{cos(x)*sin(pi/2)}}}.    (2)


Take into account that   {{{cos(pi/2)}}} = 0,   {{{sin(pi/2)}}} = 1.

Then you can continue the line (2) in this way


         {{{sin(x + pi/2)}}} = {{{sin(x)*cos(pi/2)}}} + {{{cos(x)*sin(pi/2)}}} = sin(x)*0 + cos(x)*1 = cos(x).


Thus the identity  {{{sin(x+pi/2)}}} = =cos x  is proved for all angles  x.
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Solved, explained and completed.