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Proving trigonometric identities in converse

  1. Feb 4, 2014 #1
    1. The problem statement, all variables and given/known data
    If ##\sec x-\csc x=\pm p##, show that
    ##p^{2} \sin^2 2x +4\sin 2x-4=0##

    Show conversely that if ##p^{2} \sin^2 2x +4\sin 2x-4=0##, then ##\sec x-\csc x## is equal to +p and -p.

    Find, to the nearest minute, the two values of x in the range of 0 to 360 degrees, the equation
    ##\sec x-\csc x=(1/2) (√5)##

    2. Relevant equations
    Double angle formula

    3. The attempt at a solution
    I have proved the above formula by squaring both sides and rewriting them as sines and cosines. Now how do I prove it backwards? I tried dividing both sides by ##\sin^2 2x## and it doesn't really help much.
     
  2. jcsd
  3. Feb 5, 2014 #2

    haruspex

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    Have you tried simply reversing the steps of your first proof? What goes wrong? Please post your working as far as you get.
     
  4. Feb 5, 2014 #3
    I did worked until here:
    ##\frac{4}{\sin 2x} (\frac{1}{\sin 2x}-1)=p^{2}##

    and used the double angle formula, but I got this:
    ##\csc^2 x \sec^2 x - 2\csc x \sec x=p^{2}##
     
  5. Feb 5, 2014 #4

    haruspex

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    Right so far. If you compare that with the equation you're trying to get to, what equality do you need to make them the same?
     
  6. Feb 5, 2014 #5
    I have managed to proved that conversely. :D
    Now I need to solve for x for that equation... When I substituted p=1/2 sqrt(5) into the equation, I have
    ##5\sin^2 2x+16 \sin 2x-16=0##

    Then I finally arrived at ##\sin 2x=\frac{4}{5}##. The problem is I have 4 solutions, instead of 2. :confused:
     
    Last edited: Feb 5, 2014
  7. Feb 5, 2014 #6

    AlephZero

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    That's right. From the second part, you proved the solutions of the quadratic equation satisfy sec x - csc x = plus or minus p.

    So you have to check which solutions are for +p and which are for -p.
     
  8. Feb 5, 2014 #7
    So all I have to do is to substitute them back?
     
  9. Feb 7, 2014 #8
    So all I have to do is to substitute them back?
     
  10. Feb 7, 2014 #9

    haruspex

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    Yes. Always a good check on the answer anyway.
     
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