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Math Help - Double angle trigonometry

  1. #1
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    Unhappy Double angle trigonometry....NEED HELP PLEASE!!

    Hi guys, I am new to this forum and really need some help.
    I have been trying to complete a maths assignment for my college work but have come across a question that I am totally stuck on and my lecturer wont help me!!

    Can someone help me please?!?!

    The question is:

    Give the exact value for the following:

    2tan75°
    ______
    1-tanē75°


    If someone could please please help me..
    I need to know the working and each step that was taken to get to the final answer....

    Please explain

    Thanks
    Last edited by maximus; March 21st 2013 at 11:51 AM.
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  2. #2
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    Re: Double angle trigonometry....NEED HELP PLEASE!!

    If you write the formula out using sine and cosine equivalents you get:

     \frac {2 \tan (75)}{1 - \tan ^2 (75)} = \frac { \frac {2\sin (75)}{\cos (75)}}{1 - \frac {sin^2 (75)}{\cos^2 (75)}}

    Now simplify - multiply numerator and denominator by \cos^2 (75):

     \frac {2 \sin (75) \cos (75)}{\cos^2 (75) - \sin ^2 (75)}

    Now the numerator is in the form of the identity  \sin (2x) = 2 \sin x \cos x, and the denominator is like  \cos (2x) = \cos^2x - \sin^2x. Here x = 75, so the original equation is equivalent to:

     \frac {\sin (150)}{\cos (150)}

    You can know determine an exact value for this, since you know the values for both  \sin(150) and  \cos (150).
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  3. #3
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    Re: Double angle trigonometry

    Thankyou so much
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  4. #4
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    Re: Double angle trigonometry

    An alternative method is to use a compound angle formula, since 75 = 30 + 45.
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    Re: Double angle trigonometry

    I really have no idea how to use compound formulas....could you run through it please?
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  7. #7
    Super Member ebaines's Avatar
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    Re: Double angle trigonometry

    Quote Originally Posted by maximus View Post
    I really have no idea how to use compound formulas....could you run through it please?
    You should memorize the compound angle formulas for sin(a+b) and cosine(a+b). The others can be derived from these:

     \sin(a+b) = \sin(a) \cps(b)+ \cos(a) \sin(b)
     \cos(a+b) = \cos(a)\cos(b)-\sin(a)\sin(b)


    So for your problem you have

     \tan (75)  = \frac {\sin(30+45)}{\cos(30+45)} = \frac {\sin(30)\cos(45) + \cos(30) \sin(45)}{\cos(30)\cos(45) - \sin(30| \sin(45)}

    Substitute the values for  \sin(30), \ \cos(30), \ \sin45), \ \cos(45) and then it's just a matter of working the algebra.

    Alternativey - if you can remember the compound angle for tangents:

     \tan(a+b) = \frac {\tan(a) \pm \tab (b)}{1 \mp \tan \\(a) \tab (b)}

    then by inspection you can see that for your problem if you make  a = b = 75, the answer is simply  \tan(75+75) = \tan (150).
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  8. #8
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    Re: Double angle trigonometry

    I think you mean \tan{(a + b)} \equiv \frac{\tan{(a)} \pm \tan{(b)}}{1 \mp \tan{(a)}\tan{(b)}}.
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  9. #9
    Super Member ebaines's Avatar
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    Re: Double angle trigonometry

    Quote Originally Posted by Prove It View Post
    I think you mean \tan{(a + b)} \equiv \frac{\tan{(a)} \pm \tan{(b)}}{1 \mp \tan{(a)}\tan{(b)}}.
    Yes - thanks for catching my error!
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