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prove tan2x = 3tanx - help much appreciated

tan2x = 3tanx

pls help
Reply 1
Avatar for RichE
RichE
15
don't you mean solve - that isn't an identity
Reply 2
Avatar for ryan750
ryan750
OP
2
RichE
don't you mean solve - that isn't an identity


omg - im crazy - i spent ages tryin to prove this and i just went round in circles - thanks. how do i do it now then?
Reply 3
Avatar for Syncman
Syncman
is that even possible? :s-smilie:
Reply 4
Avatar for RichE
RichE
15
start with the tan2x formula for the LHS and factorise
Reply 5
Avatar for RichE
RichE
15
you've missed when tanx=0 and tanx = -1/rt(3) possibilites
Reply 6
Avatar for ryan750
ryan750
OP
2
RichE
you've missed when tanx=0 and tanx = -1/rt(3) possibilites


you are right. I was just thinking how the answers 0, pi and 2pi came from. can u explain where these answers cum from?
Reply 7
Avatar for RichE
RichE
15
ryan750
you are right. I was just thinking how the answers 0, pi and 2pi came from. can u explain where these answers cum from?


just don't divide through by tanx and take care to include positive and negative square roots. Newton's answer was otherwise perfect
Reply 8
Avatar for ryan750
ryan750
OP
2
RichE
just don't divide through by tanx and take care to include positive and negative square roots. Newton's answer was otherwise perfect


Where is newtons answer?
Reply 9
Avatar for RichE
RichE
15
he seems to have retracted it
Reply 10
Avatar for ryan750
ryan750
OP
2
RichE
he seems to have retracted it


im completely stuck - please can u show me how to get the 0, pi and 2pi answers. i dont see how u cant not divide thru by tanx? please help
Reply 11
Avatar for ryan750
ryan750
OP
2
ryan750
im completely stuck - please can u show me how to get the 0, pi and 2pi answers. i dont see how u cant not divide thru by tanx? please help


i have rep availbale for the one that shows me the way to get to 0, pi and 2pi.
Reply 12
Avatar for Newton
Newton
tan2x=3tanx

=>(2tanx/(1-((tanx)^2))=3tanx

=>2tanx=3tanx(1-((tanx)^2))

=>2tanx=3tanx-3((tanx)^3)

=>3((tanx)^3)-tanx=0

=>tanx(3((tanx)^2)-1)=0

=>tanx=0 OR tanx=(1/sqrt(3)) OR tanx=-(1/sqrt(3))

tanx=0=>x=arctan(0)=0, (Pi), 2(Pi)

tanx=(1/sqrt(3))=>x=arctan(1/sqrt(3))=((Pi)/6), (7/6)(Pi)

tanx=arctan(-(1/sqrt(3)))=-arctan((1/sqrt(3)))

Supposing that the range, 0<=R<=(2(Pi)), tanx=-(1/sqrt(3))) does not give a valid range of values, so:

x=0, ((Pi)/6), (Pi), (7/6)(Pi), 2(Pi)

Newton.

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