How do you solve # sin (x- (pi/2))= cos x#?

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Answer 1 · 2016-01-27T08:45:14

#x=pi/2#

#sin A=cos (pi/2-A)#

Therefore, from the given equation

#sin (x-pi/2)=cos x#

If #A=x-pi/2#, then

#sin (x-pi/2)=cos (pi/2-(x-pi/2))# it follows that

#cos x=cos (pi/2-(x-pi/2))#

Equate the angles now

#x=pi/2-(x-pi/2))#

#x=pi/2-x+pi/2#

#x+x=pi/2+pi/2#

#2x=pi#

#x=pi/2#

Check using the original equation:

#sin (x-pi/2)=cos x#

#sin (pi/2-pi/2)=cos (pi/2)#

#sin 0=cos (pi/2)#

#0=0#

correct, therefore #x=pi/2#