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Thread: trigonometry Addition Formulae (ii)

  1. #1
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    trigonometry Addition Formulae (ii)

    As you may see, I am struggling at this.



    Show that, for triangle RST, t = r sin (alpha + beta) / sin alpha



    iv tried using the sine rule,
    t / sin(180 - alpha - beta) = r / sin alpha
    t = r / sin alpha x sin(180 - alpha - beta)

    After that, i get confused because usually i would expand the sin bracket. ie:
    sin(45+45) = sin 45 cos 45 + cos 45 sin 45

    but in this case, i have 180, alpha, and beta in one bracket.......

    Am i even taking the right steps?
    Please help.
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  2. #2
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    Hello, r_maths!

    Show that, for $\displaystyle \Delta\!:\;\;t \:= \:\frac{r\sin(\alpha + \beta)}{\sin\alpha}$
    Code:
                  T
                  *
                 *  *
                *     *
             s *        * r
              *           *
             *              *
            *                 *
           * α                β *
        R * * * * * * * * * * * * * S
                      t

    From the Law of Sines, we have: .$\displaystyle \frac{t}{\sin T} \:=\:\frac{r}{\sin\alpha}$ [1]

    Note that: .$\displaystyle T \;=\;180^o - (\alpha + \beta)$
    . . and: .$\displaystyle \sin T \;= \;\sin[180^o - (\alpha + \beta)] \;=\;\sin(\alpha + \beta)$

    So [1] becomes: .$\displaystyle \frac{t}{\sin(\alpha + \beta)} \:=\:\frac{r}{\sin\alpha}\quad\Rightarrow\quad t \:=\:\frac{r\sin(\alpha + \beta)}{\sin\alpha} $

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  3. #3
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    Quote Originally Posted by Soroban View Post
    Hello, r_maths!
    So [1] becomes: .$\displaystyle \frac{t}{\sin(\alpha + \beta)} \:=\:\frac{r}{\sin\alpha}\quad\Rightarrow\quad t \:=\:\frac{r\sin(\alpha + \beta)}{\sin\alpha} $

    [/size]
    $\displaystyle \frac{t}{\sin(\alpha + \beta)} $

    I know your right but,
    SinT = sin[180-(alpha+beta) = sin (alpha + beta)

    Where did the 180 go? (is it sin 180 = 0)?
    and the minus sign? wouldnt that make it sin(-aplha - beta)
    :S

    thanks for your help, i really appreciate.
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  4. #4
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    Nevermind, looked back at my work, worked out why.
    Thanks!!
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  5. #5
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    Just for completeness:
    $\displaystyle sin(180 - [ \alpha + \beta]) = sin(180) \cdot cos(\alpha + \beta) - sin(\alpha + \beta) \cdot cos(180)$

    Since $\displaystyle sin(180) = 0$ and $\displaystyle cos(180) = -1$ we get
    $\displaystyle sin(180 - [ \alpha + \beta]) = 0 - - sin(\alpha + \beta) = sin(\alpha + \beta)$

    -Dan
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