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Homework Help: Inverse tangent proof

  1. Oct 18, 2006 #1
    Show that [tex]\arctan{x} + \arctan {y} = \arctan { \frac{x+y}{1-xy} }[/tex] when [tex]x = \frac{1}{2}\ and \y = \frac{1}{3}[/tex] but not when [tex]x = 2\ and \y = 3[/tex]

    I've tried taking the tangent of both sides but I don't know what to do then when I've got [tex]\tan ( \arctan{x} + \arctan{y} ) = \frac{x+y}{1-xy}[/tex]

    Any help would be greatly appreciated. Thanks!
     
    Last edited: Oct 18, 2006
  2. jcsd
  3. Oct 18, 2006 #2
    Use the fact that [tex] \tan(u+v) = \frac{\tan u + \tan v}{1-\tan u \tan v} [/tex]
     
  4. Oct 19, 2006 #3

    HallsofIvy

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    I don't see what the difficulty is. You are aske to show that this equation is satisfied when [itex]x= \frac{1}{2}[/itex] and [itex]y= \frac{1}{3}[/itex] but not when x= 2 and y= 3.

    Okay, plug those values in and evaluate. Courtrigrax's method would appear to be a general method of proving that it is an identity: true for all x and y which contradicts the problem!
     
  5. Oct 19, 2006 #4

    HallsofIvy

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    I'm not clear what the problem is. You are asked to show that this equation is satisfied when [itex]x= \frac{1}{2}[/itex] and [itex]y= \frac{1}{3}[/itex] but not when x= 2 and y= 3.

    Okay, plug those values in and evaluate.

    Courtrigrax's method would appear to be a general method of proving that it is an identity: true for all x and y, which contradicts the statement of the problem!
     
  6. Oct 19, 2006 #5
    Ok, so I used the identity [tex] \tan(u+v) = \frac{\tan u + \tan v}{1-\tan u \tan v} [/tex] but all it does is bring me to the equation [tex]\frac{x+y}{1-xy} = \frac{x+y}{1-xy}[/tex]. I had tried substituting the values in before and found that the statement held true for both sets of x and y, but I assumed I must have done something wrong. I told my teacher and she said I was wrong :confused:

    So have I messed up? Or does the equation hold true for x = 2 and y = 3 thus rendering the proof impossible?
     
  7. Oct 19, 2006 #6
    The equation does not hold for x = 2 and y = 3. Just plug in the values. Also, can you see that it will work for [tex] x > -1 [/tex] , [tex] y < 1 [/tex]? Why is this?

    It is because the left hand side is not defined if [tex] xy = 1 [/tex]
     
    Last edited: Oct 19, 2006
  8. Oct 20, 2006 #7
    I plugged in the values and got the following...

    [tex]\arctan {2}\ + \arctan {3} = \arctan {-1}[/tex]
    [tex]2.35619... = -0.78539...[/tex]

    However another solution to [tex]\arctan {-1}[/tex] is [tex]-0.78539... + pi = 2.35619...[/tex] which makes the statement hold true :confused: Am I not allowed to take anything other than the principal value? If so, why not?

    Also, you said that [tex] x > -1 [/tex], [tex] y < 1 [/tex]. Are you saying that these are the only values of x and y that the equation will hold for? That doesn't seem to make sense, as if, for example, I use x = -5 and y = 10 the equation will still hold true.
     
    Last edited: Oct 20, 2006
  9. Oct 20, 2006 #8

    HallsofIvy

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    tangent(x) has a discontinuity at [itex]\pi/2[/itex] or approximately 1.57.
     
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