Solve cos2A=sqrt(2)(cosA-sinA)cos2A=√2(cosA−sinA)?

2 Answers
Sunayan S
Oct 21, 2017

See the answer below...

Explanation:

cos2A=sqrt2(cosA-sinA)cos2A=√2(cosA−sinA)
=>cos2A(cosA+sinA)=sqrt2(cos^2A-sin^2A)⇒cos2A(cosA+sinA)=√2(cos2A−sin2A)
=>cos2A(cosA+sinA)=sqrt2 cdot cos2A⇒cos2A(cosA+sinA)=√2⋅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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