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Math Help - Deriving a trig formula

  1. #1
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    Deriving a trig formula

    I'm practicing trig and so I'm working through some exercises. I'm having some trouble figuring this one out:

    tan(\alpha + \beta) = \frac{tan\, \alpha + tan\, \beta}{1 - tan\, \alpha\, tan\, \beta}

    First I've rewritten tan(\alpha + \beta) as \frac{sin(\alpha + \beta) }{cos(\alpha + \beta)} which equates to:
    \frac{sin\, \alpha\, cos\, \beta + \cos\,\alpha\,sin\beta}{cos\,\alpha\,cos\beta - sin\,\alpha\,sin\,\beta}.

    Then I'm kind of stuck, so I tried going the other way around:

    \frac{tan\,\alpha + tan\,\beta}{1 - tan\,\alpha \, tan\, \beta} =

    \frac{\frac{sin\,\alpha}{cos\,\alpha}+ \frac{sin\alpha\,\beta }{cos\,\beta}}{1 - \frac{sin\, \alpha}{cos\, \alpha}\frac{sin\, \beta}{cos\, \beta}}

    Then I multiply the 1/1 in the denominator by cos a cos b, to make them equal, to get rid of the 1, then I rewrite it according to \frac{\frac{a}{b}}{\frac{c}{d}} = \frac{ad}{bc} which leaves me with a lot of sin and cos, but stuck as well. Some help would be nice
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  2. #2
    MHF Contributor Siron's Avatar
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    Re: Deriving a trig formula

    Quote Originally Posted by Lepzed View Post
    I'm practicing trig and so I'm working through some exercises. I'm having some trouble figuring this one out:

    tan(\alpha + \beta) = \frac{tan\, \alpha + tan\, \beta}{1 - tan\, \alpha\, tan\, \beta}

    First I've rewritten tan(\alpha + \beta) as \frac{sin(\alpha + \beta) }{cos(\alpha + \beta)} which equates to:
    \frac{sin\, \alpha\, cos\, \beta + \cos\,\alpha\,sin\beta}{cos\,\alpha\,cos\beta - sin\,\alpha\,sin\,\beta}.
    Divide every term by \cos(\alpha)\cos(\beta).
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  3. #3
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    Re: Deriving a trig formula

    Aah, got it now yeah! Thanks!

    Right, so I know HOW to solve it, but how could I get to sin a cos b? Just try?
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  4. #4
    MHF Contributor Siron's Avatar
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    Re: Deriving a trig formula

    Do you mean 'How could I get to cos(a)cos(b)'? Or? ...
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  5. #5
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    Re: Deriving a trig formula

    Yeah, well, you know how to solve this, so for you it's obvious to divide by cos(a)cos(b). For me it's not, now I can just go about trying some stuff till I find something that eventually would work, but that doesn't seem really productive or educated.
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  6. #6
    Super Member Quacky's Avatar
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    Re: Deriving a trig formula

    You can just make an educated guess; you have a \sin\alpha\cos\beta as the first term in the numerator, and you need to have a \tan\alpha which implies the division, and you can then verify it by checking with each other term.
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