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Proving trigo identity

  1. May 14, 2005 #1
    Prove that
    [tex]\sec x + \tan x = \tan \left (\frac{\pi}{4} + \frac{x}{2}\right )[/tex]

    I've got to
    [tex]\sec x + \tan x = \frac{1+\sin x}{\cos x}[/tex]
    and then I was stuck. Tried half angle but it didn't seem to work.

    Help please.
     
  2. jcsd
  3. May 14, 2005 #2

    Hurkyl

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    There are two sides to the equation -- it sounds like you've only fiddled with the left hand side. :frown:
     
  4. May 14, 2005 #3

    arildno

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    Neat!
    I've never seen that trig. identity before..
     
  5. May 14, 2005 #4
    Well, manipulating RHS gives
    [tex]\tan \left (\frac{\pi}{4} + \frac{x}{2} \right )[/tex]
    [tex]= \frac{\tan\frac{\pi}{4}+\tan\frac{x}{2}}{1-\tan\frac{\pi}{4}\tan\frac{x}{2}}[/tex]
    [tex]=\frac{1+\tan\frac{x}{2}}{1-\tan\frac{x}{2}}[/tex]

    and applying half angle to LHS gives
    [tex]\frac{1+\sin x}{\cos x}[/tex]
    [tex]=\frac{1+2\sin\frac{x}{2}\cos\frac{x}{2}}{1-2\sin\frac{x}{2}\sin\frac{x}{2}}[/tex]

    Hmm.. how can [tex]\tan\frac{x}{2}[/tex] be equal to
    [tex]2\sin\frac{x}{2}\cos\frac{x}{2}}[/tex]
    and
    [tex]2\sin\frac{x}{2}\sin\frac{x}{2}}[/tex] at the same time?

    Any help please..
     
  6. May 14, 2005 #5

    Curious3141

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    Here's help. :)

    Try to manipulate the more complicated side to get the less complicated side. In this case, work on the RHS to get the LHS. Try not to work from both sides at once.

    Let the respective sin, cos and tan trig ratios of x/2 be s, c and t. Let those of x be S, C and T. I'm doing this because I'm really fed up of clunky LaTex.

    Taking it from where you left off,

    RHS :

    [tex]\frac{1+\tan\frac{x}{2}}{1-\tan\frac{x}{2}}[/tex]
    [tex]=\frac{(1 + t)^2}{(1-t)(1 + t)}[/tex]
    [tex]=\frac{1 + t^2 + 2t}{1 - t^2}[/tex]
    [tex]=\frac{1 + t^2 + 2t}{\frac{c^2 - s^2}{c^2}}[/tex]
    [tex]=\frac{(1 + t^2 + 2t)(c^2)}{C}[/tex]
    [tex]=\frac{c^2 + s^2 + 2sc}{C}[/tex]
    [tex]=\frac{1 + S}{C}[/tex]
    [tex]=\sec{x} + \tan{x}[/tex] (QED)
     
  7. May 14, 2005 #6

    Curious3141

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    BTW, the proof (and the original identity) fail for t = -1. In the proof, it's because I multiply the RHS by (1+t)/(1+t). In the orig. identity, the LHS becomes undefined while the RHS remains finite (so the failure is consistent).
     
  8. May 14, 2005 #7
    Thanks for your help :smile:
     
  9. May 15, 2005 #8

    dextercioby

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    You know

    [tex] \sec x+\tan x=\frac{1}{\cos x}+\frac{\sin x}{\cos x}=\frac{1+\sin x}{\cos x} [/tex] (1)

    Th RHS is

    [tex] \frac{1+\tan\frac{x}{2}}{1-\tan\frac{x}{2}}=\frac{1+\frac{\sin^{2}\frac{x}{2}}{\cos^{2}\frac{x}{2}}+2\frac{\sin\frac{x}{2}}{\cos\frac{x}{2}}}{1-\frac{\sin^{2}\frac{x}{2}}{\cos^{2}\frac{x}{2}}} [/tex]

    [tex]=\frac{\cos^{2}\frac{x}{2}+\sin^{2}\frac{x}{2}+2\cos\frac{x}{2}\sin\frac{x}{2}}{\cos^{2}\frac{x}{2} -\sin^{2}\frac{x}{2}}=\frac{1+\sin x}{\cos x} [/tex]

    (Q.e.d.)

    ,pretty simple,right...?

    Daniel.
     
    Last edited: May 15, 2005
  10. May 15, 2005 #9

    arildno

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    I think he mentioned that identity in post 1, Daniel..:wink:
     
  11. May 15, 2005 #10

    dextercioby

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    Hehe,i thought he went backwards starting with the RHS.:tongue2:

    Daniel.
     
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