How do you find the value of #theta# for which #cos(theta+pi/2)# does not equal #costheta+cos(pi/2)#?

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Answer 1 · 2017-01-20T08:52:01

See explanation.

The inequality given is:

#cos(x+pi/2)!=cosx+cos(pi/2)#

This inequality can be transformed into:

#cos(x+pi/2)!=cosx# because #cos(pi/2)=0#

Now we can move #cosx# left:

#cos(x+pi/2)-cosx!=0#

Now we use the formula

#cosx-cosy=-2sin((x+y)/2)*sin((x-y)/2)#

#-2sin(x+pi/4)sin(pi/4)!=0#

#sin(x+pi/4)!=0#

This inequality is fulfilled for:

#x+pi/4 !=k*pi# for #k in ZZ#

#x !=-pi/4+k*pi# for #k in ZZ#