Solve #cos2A=sqrt(2)(cosA-sinA)#?

2 Answers
Sunayan S
Oct 21, 2017

See the answer below...

Explanation:

#cos2A=sqrt2(cosA-sinA)#
#=>cos2A(cosA+sinA)=sqrt2(cos^2A-sin^2A)#
#=>cos2A(cosA+sinA)=sqrt2 cdot cos2A#
#=>cancel(cos2A)(cosA+sinA)=sqrt2 cdot cancel(cos2A#
#=>(cosA+sinA)=sqrt2#
#=>sin^2A+cos^2A+2sinAcosA=2#[squared both side]
#=>1+sin2A=2#
#=>sin2A=1=sin90^@#
#=>2A=90^@#
#=>A=45^@#

HOPE THE ANSWER HELPS...

THANK YOU...

P dilip_k
Nov 9, 2017

#cos2A=sqrt2(cosA-sinA)#

#=>cos^2A-sin^2A-sqrt2(cosA-sinA)=0#

#=>(cosA-sinA)(cosA+sinA)-sqrt2(cosA-sinA)=0#

#=>(cosA-sinA)(cosA+sinA-sqrt2)=0#

When

#cosA+sinA=0#

#=>tanA=1=tan(pi/4)#

#=>A=npi+pi/4" where " n in ZZ#

#cosA+sinA=sqrt2#

#=>1/sqrt2cosA+1/sqrt2sinA=1#

#=>cos(pi/4)cosA+sin(pi/4)sinA=1#

#=>cos(A-pi/4)=1#

#=>A=2mpi+pi/4" where " m in ZZ#

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