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anonymous

  • one year ago

Prove

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  1. anonymous
    • one year ago
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    \[\frac{ sinx+1 }{ cosx}=\frac{ sinx-cosx+1 }{ sinx+cosx-1 }\] @data_lg2

  2. nincompoop
    • one year ago
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    wow

  3. zzr0ck3r
    • one year ago
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    So wolfram puts a bunch of restrictions on the solution set, but it is true:)

  4. mathstudent55
    • one year ago
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    \(\dfrac{ \sin x+1 }{ \cos x}=\dfrac{ \sin x-\cos x+1 }{ \sin x+\cos x-1 }\) \((\sin x+1)(\sin x+\cos x-1) = \cos x (\sin x-\cos x+1)\) \(\sin^2 x + \sin x \cos x - \sin x + \sin x + \cos x -1 = \cos x \sin x - \cos^2 x + \cos x\) \(\sin^2 x + \cancel{\sin x \cos x} -\cancel{ \sin x} + \cancel{\sin x} + \cancel{\cos x} -1 = \cancel{\cos x \sin x} - \cos^2 x + \cancel{\cos x}\) \(\sin^2 x - 1 = - \cos^2 x\) \(\sin^2x + \cos^2 x = 1\)

  5. anonymous
    • one year ago
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    An efficient/proper way would be to multiply LHS by (sinx+cosx-1) and simplify sin^2x as 1-cos^2x to get RHS

  6. anonymous
    • one year ago
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    I'm sorry @mathstudent55 but that's not proving. Proving means find the expression from the right hand using just the expression from the left hand. To achieve this, you can multiply and divide any expression you think it's convenient without actually changing the original expression, so you will likely have to multiply by 1 or 2/2 or sinx/sinx, etc. My answer is quite long, but if you wait, i'll finish typing it in latex

  7. anonymous
    • one year ago
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    @RicardoDelfin It is a method of proving for sure. But not the proper/efficient way. In-fact not the method that teachers would allow you to use. But me and bunch of colleagues did a research on this specific topic and most of the Grad level professors accepted it as another way of proving an equation. You cannot simply say that it is not proving.

  8. anonymous
    • one year ago
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    \[\frac{ \sin x +1 }{ \cos x } \left( \frac{ 2 \sin x }{ 2 \sin x } \right) = \frac{ 2\sin^2x + 2 \sin x }{ 2 \sin x \cos x }\] Remember that \[(\sin x + \cos x)^2 =1 + 2 \sin x \cos x\] So in our expression \[\frac{ 2\sin^2x + 2 \sin x }{ (\sin x + \cos x)^2-1 }=\frac{ 2\sin^2x + 2 \sin x }{ (\sin x + \cos x -1)(\sin x+\cos x +1) }\] so if we want to get rid of sinx +cosx +1 we need to multiply our expression again by another quotient i'll pick sinx-cosx +1 this time \[\frac{ 2\sin^2x + 2 \sin x }{ (\sin x + \cos x -1)(\sin x +\cos x +1) }(\frac{ \sin x - \cos x +1 }{ \sin x - \cos x +1 })\] \[\frac{ (2\sin^2x + 2 \sin x)(\sin x -\cos x +1) }{ (\sin x + \cos x -1)(\sin x - cos x +1)(\sin x + \cos x +1) }\] Solving this \[(\sin x + \cos x -1)(\sin x - \cos x +1)=2\sin^2x +2\sin x\] so our expression would be\[\frac{ (2\sin^2x +2 \sin x)(\sin x - \cos x +1) }{ (2\sin^2x + 2 \sin x)(\sin x - \cos x +1) }=\frac{ \sin x -\cos x +1 }{ \sin x - \cos x +1 }\] which is what we wanted to prove

  9. anonymous
    • one year ago
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    @d3v13 I see, well you have a point, but like you said in most of schools that method is not considered as proper demonstration. Anyway i didnt want to be mean, just wanted to point out that the method he used is not how usually proving problems are conventionally solved

  10. anonymous
    • one year ago
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    i made a typo on the last line, and dont know how to edit (lol) it should be \[\frac{ (2\sin^2x + 2 \sin x)(\sin x - \cos x +1) }{ (2\sin^2x + 2 \sin x)(\sin x + \cos x +1) }=\frac{ \sin x - \cos x + 1 }{ \sin x + \cos x +1 }\]

  11. zzr0ck3r
    • one year ago
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    You of course can't start with what you are trying to prove....you can of course use this as side work and then you may start with something you know is true, and through logical implication arrive at what you are proving.

  12. zzr0ck3r
    • one year ago
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    @d3v13 how in the world can something be a correct proof but not proper? That alone says that the proof was not correct. This is crazy talk up in here....

  13. anonymous
    • one year ago
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    Dear @zzr0ck3r ; Correct and proper have two different meaning, isn't that enough to be something which is correct but not proper. If you look closely at both the solutions here, it's the same, but done in two different ways.

  14. zzr0ck3r
    • one year ago
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    What then did you mean my not proper? So you are arguing that when proving something, you can start with the thing you are proving? Just say that out loud.

  15. zzr0ck3r
    • one year ago
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    by*

  16. anonymous
    • one year ago
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    Ofcuase you can. For example in Tropical mathematics, let's just say that you want to prove 1=1. I cannot think of a method to start proving this without using 1=1. Please be my guest if you know any.

  17. anonymous
    • one year ago
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    @myjopji

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