anonymous
  • anonymous
True or False? tan^2x= 1+ cos2x / 1-cos2x
Mathematics
chestercat
  • chestercat
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SolomonZelman
  • SolomonZelman
\(\large\color{black}{ \displaystyle \tan^2(x)=\frac{ 1+\cos^2(x) }{1-\cos^2(x)} }\)
SolomonZelman
  • SolomonZelman
verify or prove ?
P0sitr0n
  • P0sitr0n
Multiply both sides by 1-cos^2 x and simplify , keeping in mind that \[tan(x)=\frac{sin(x)}{cos(x)}\]

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anonymous
  • anonymous
It does not specify on the question itself but I would assume proving.
SolomonZelman
  • SolomonZelman
(verify = manipulate 1 side at a time only prove= both sides can be manipulate) this is a traditional interpretation/definition of these words.
SolomonZelman
  • SolomonZelman
ok, lets just in case verify it. will play 1 side at a time
anonymous
  • anonymous
Ok
SolomonZelman
  • SolomonZelman
\(\large\color{black}{ \displaystyle \tan^2(x)=\frac{ 1+\cos^2(x) }{1-\cos^2(x)} }\) can you re-write the bottom ?
SolomonZelman
  • SolomonZelman
use the fact that \(\large\color{black}{ \displaystyle \sin^2(x)+\cos^2(x)=1 }\)
anonymous
  • anonymous
So you would be able to re-write the bottem
SolomonZelman
  • SolomonZelman
yes, in the fact I just offered, subtract cos^2(x) from both sides, and tell me what you get
anonymous
  • anonymous
From the sin^2(x) one?
SolomonZelman
  • SolomonZelman
yes, from the sin^2(x)+cos^2(x)=1, there subtract cos^2(x) from both sides
anonymous
  • anonymous
\[\sin ^{2}(x)=-\cos ^{2}(x)+1\]
SolomonZelman
  • SolomonZelman
yes sin^2(x)=1-cos^2(x)
SolomonZelman
  • SolomonZelman
now, we have \(\large\color{black}{ \displaystyle \tan^2(x)=\frac{ 1+\cos^2(x) }{1-\cos^2(x)} }\) can you re-write the bottom for me?
Pawanyadav
  • Pawanyadav
That is false As 1+cos2x=1+2cos^2x-1 =2cos^2x And 1-cos2x=1-(1-2sin^2x) =2sin^2x Put these values you will get the answer.
Loser66
  • Loser66
just put x = pi/4 , you can see how it is false
anonymous
  • anonymous
When I rearrange this @SolomonZelman, will I be taking the -cos^2(x) away from the denominator and onto the numerator? I am not very good at trig
SolomonZelman
  • SolomonZelman
\(\large\color{black}{ \displaystyle \tan^2(x)=\frac{ 1+\cos^2(x) }{1-\cos^2(x)} }\) \(\large\color{black}{ \displaystyle \tan^2(x)=\frac{ 1+\cos^2(x) }{\sin^2(x)} }\) that is what I meant
SolomonZelman
  • SolomonZelman
anyway, i think plugging in value to disprove is valid enough.
SolomonZelman
  • SolomonZelman
unless u want to go and do it without plug ins
anonymous
  • anonymous
Would I just have to plug in pi/4 to the equation?
Pawanyadav
  • Pawanyadav
But that's not a proper method
anonymous
  • anonymous
Oh ok
anonymous
  • anonymous
Well thank you guys for the help and time
Loser66
  • Loser66
Who said : proving the invalid of an expression by a counterexample is not a proper method?
Loser66
  • Loser66
That is the MOST proper method to prove something wrong.
SolomonZelman
  • SolomonZelman
\(\large\color{black}{ \displaystyle \tan^2(x)=\frac{ 1+\cos^2(x) }{1-\cos^2(x)} }\) \(\large\color{black}{ \displaystyle \frac{\sin^2x}{\cos^2x}=\frac{ (1+\cos^2x) }{(1-\cos^2x)} }\) \(\large\color{black}{ \displaystyle \frac{\sin^2x}{\cos^2x}=\frac{ (1+\cos^2x) }{(1-\cos^2x)} }\) \(\large\color{black}{ \displaystyle \sin^2x(1-\cos^2x)=\cos^2x(1+\cos^2x) }\) \(\large\color{black}{ \displaystyle \sin^2x-\sin^2x\cos^2x=\cos^2x+\cos^2x\cos^2x}\) \(\large\color{black}{ \displaystyle \sin^2x=\sin^2x\cos^2x+\cos^2x+\cos^2x\cos^2x}\) \(\large\color{black}{ \displaystyle \sin^2x=\cos^2x(\sin^2x+1+\cos^2x)}\) \(\large\color{black}{ \displaystyle \sin^2x=\cos^2x(2)}\) \(\large\color{black}{ \displaystyle \sin^2x=2(1- \sin^2x)}\) \(\large\color{black}{ \displaystyle \sin^2x=2- 2\sin^2x}\) \(\large\color{black}{ \displaystyle 3\sin^2x=2}\) \(\large\color{black}{ \displaystyle \sin^2x=2/3}\) \(\large\color{black}{ \displaystyle \sin x=\pm\sqrt{2/3}}\) \(\large\color{black}{ \displaystyle x=\sin^{-1}\left(\pm\sqrt{2/3}\right)}\)
SolomonZelman
  • SolomonZelman
if your definition were to be true for all numbers, then this solution for x would be such that includes all numbers. \(\large\color{black}{ \displaystyle \sin^{-1}\left(\pm\sqrt{2/3}\right)}\) is no way a representation of all [real] numbers!
anonymous
  • anonymous
*Takes a picture for future reference* Thank you for the explanation! This will help me 200x better than before!
SolomonZelman
  • SolomonZelman
i took everything down piece by piece. it isn't that bad usually though

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