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anonymous

  • one year ago

True or False? tan^2x= 1+ cos2x / 1-cos2x

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  1. SolomonZelman
    • one year ago
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    \(\large\color{black}{ \displaystyle \tan^2(x)=\frac{ 1+\cos^2(x) }{1-\cos^2(x)} }\)

  2. SolomonZelman
    • one year ago
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    verify or prove ?

  3. P0sitr0n
    • one year ago
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    Multiply both sides by 1-cos^2 x and simplify , keeping in mind that \[tan(x)=\frac{sin(x)}{cos(x)}\]

  4. anonymous
    • one year ago
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    It does not specify on the question itself but I would assume proving.

  5. SolomonZelman
    • one year ago
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    (verify = manipulate 1 side at a time only prove= both sides can be manipulate) this is a traditional interpretation/definition of these words.

  6. SolomonZelman
    • one year ago
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    ok, lets just in case verify it. will play 1 side at a time

  7. anonymous
    • one year ago
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    Ok

  8. SolomonZelman
    • one year ago
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    \(\large\color{black}{ \displaystyle \tan^2(x)=\frac{ 1+\cos^2(x) }{1-\cos^2(x)} }\) can you re-write the bottom ?

  9. SolomonZelman
    • one year ago
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    use the fact that \(\large\color{black}{ \displaystyle \sin^2(x)+\cos^2(x)=1 }\)

  10. anonymous
    • one year ago
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    So you would be able to re-write the bottem

  11. SolomonZelman
    • one year ago
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    yes, in the fact I just offered, subtract cos^2(x) from both sides, and tell me what you get

  12. anonymous
    • one year ago
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    From the sin^2(x) one?

  13. SolomonZelman
    • one year ago
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    yes, from the sin^2(x)+cos^2(x)=1, there subtract cos^2(x) from both sides

  14. anonymous
    • one year ago
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    \[\sin ^{2}(x)=-\cos ^{2}(x)+1\]

  15. SolomonZelman
    • one year ago
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    yes sin^2(x)=1-cos^2(x)

  16. SolomonZelman
    • one year ago
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    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?

  17. Pawanyadav
    • one year ago
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    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.

  18. Loser66
    • one year ago
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    just put x = pi/4 , you can see how it is false

  19. anonymous
    • one year ago
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    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

  20. SolomonZelman
    • one year ago
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    \(\large\color{black}{ \displaystyle \tan^2(x)=\frac{ 1+\cos^2(x) }{1-\cos^2(x)} }\) \(\large\color{black}{ \displaystyle \tan^2(x)=\frac{ 1+\cos^2(x) }{\sin^2(x)} }\) that is what I meant

  21. SolomonZelman
    • one year ago
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    anyway, i think plugging in value to disprove is valid enough.

  22. SolomonZelman
    • one year ago
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    unless u want to go and do it without plug ins

  23. anonymous
    • one year ago
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    Would I just have to plug in pi/4 to the equation?

  24. Pawanyadav
    • one year ago
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    But that's not a proper method

  25. anonymous
    • one year ago
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    Oh ok

  26. anonymous
    • one year ago
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    Well thank you guys for the help and time

  27. Loser66
    • one year ago
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    Who said : proving the invalid of an expression by a counterexample is not a proper method?

  28. Loser66
    • one year ago
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    That is the MOST proper method to prove something wrong.

  29. SolomonZelman
    • one year ago
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    \(\large\color{black}{ \displaystyle \tan^2(x)=\frac{ 1+\cos^2(x) }{1-\cos^2(x)} }\) \(\large\color{black}{ \displaystyle \frac{\sin^2x}{\cos^2x}=\frac{ (1+\cos^2x) }{(1-\cos^2x)} }\) \(\large\color{black}{ \displaystyle \frac{\sin^2x}{\cos^2x}=\frac{ (1+\cos^2x) }{(1-\cos^2x)} }\) \(\large\color{black}{ \displaystyle \sin^2x(1-\cos^2x)=\cos^2x(1+\cos^2x) }\) \(\large\color{black}{ \displaystyle \sin^2x-\sin^2x\cos^2x=\cos^2x+\cos^2x\cos^2x}\) \(\large\color{black}{ \displaystyle \sin^2x=\sin^2x\cos^2x+\cos^2x+\cos^2x\cos^2x}\) \(\large\color{black}{ \displaystyle \sin^2x=\cos^2x(\sin^2x+1+\cos^2x)}\) \(\large\color{black}{ \displaystyle \sin^2x=\cos^2x(2)}\) \(\large\color{black}{ \displaystyle \sin^2x=2(1- \sin^2x)}\) \(\large\color{black}{ \displaystyle \sin^2x=2- 2\sin^2x}\) \(\large\color{black}{ \displaystyle 3\sin^2x=2}\) \(\large\color{black}{ \displaystyle \sin^2x=2/3}\) \(\large\color{black}{ \displaystyle \sin x=\pm\sqrt{2/3}}\) \(\large\color{black}{ \displaystyle x=\sin^{-1}\left(\pm\sqrt{2/3}\right)}\)

  30. SolomonZelman
    • one year ago
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    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!

  31. anonymous
    • one year ago
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    *Takes a picture for future reference* Thank you for the explanation! This will help me 200x better than before!

  32. SolomonZelman
    • one year ago
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    i took everything down piece by piece. it isn't that bad usually though

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