How do you show $cos(x+pi/2)+cos(x-pi/2)=0$ ?

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Answer 1 · 2018-03-07T12:49:56

We need to use the trig identity:
$cos(A+-B)=cosAcosB∓sinAsinB$

Using this, we get:
$cos(x+pi/2)+cos(x-pi/2)=(cosxcos(pi/2)+sinxsin(pi/2))+(cosxcos(pi/2)-sinxsin(pi/2))$

$cos(pi/2)=0$
$sin(pi/2)=1$

$cos(x+pi/2)+cos(x-pi/2)=(0cosx+1sinx)+(0cosx-1sinx)=sinx-sinx=0$

How do you show cos(x+pi/2)+cos(x-pi/2)=0cos(x+pi/2)+cos(x-pi/2)=0?