IF #P={x:sinx-cosx=sqrt2 cosx} and Q={x:sinx+cosx=sqrt2 sinx} # then show that P=Q?

Question

3618 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-11-02T08:41:29

#P={x:sinx-cosx=sqrt2 cosx} and Q={x:sinx+cosx=sqrt2 sinx} #

For the relation

#sinx-cosx=sqrt2 cosx#

#=>1/sqrt2cosx-1/sqrt2sinx=-cosx#

#=>cos(pi/4)cosx-sin(pi/4)sinx=-cosx#

#=>cos(x+pi/4)=cos(pi-x)#

#=>x+pi/4=2npi+pi-x#

#=>x=npi+(3pi)/8" where " nin ZZ#

So #P={x:x=npi+(3pi)/8" where " nin ZZ}#

For the relation

#sinx+cosx=sqrt2 sinx#

#=>1/sqrt2cosx+1/sqrt2sinx=sinx#

#=>cos(pi/4)cosx+cos(pi/4)sinx=cos(pi/2-x)#

#=>cos(x-pi/4)=cos(pi/2-x)#

#=>x-pi/4=2npi+(pi/2-x) " where " n in ZZ#

#=>2x=2npi+(3pi)/4 " where " n in ZZ#

#=>x=npi+(3pi)/8" where " nin ZZ#

So #Q={x:x=npi+(3pi)/8" where " nin ZZ}#

Hence we have #P=Q#