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Math Help - Inverse Trig functions- Find sin(arctanp+arctan1/p) if p>0

  1. #1
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    Arrow Inverse Trig functions- Find sin(arctanp+arctan1/p) if p>0

    Hi, I was given this question for homework but I'm stuck. Determine sin(arctanp+arctan1/p) if p>0
    After using the compound angle formula and substituting functions, I have no idea what to do next:
    I started off by using the compound angle formula, giving me: [sin(arctan(p))]*[cos(arctan(1/p)] + [cos(arctan(p)]*[sin(arctan(1/p)]. From there I Let arctan(p)=a. Through manipulation I therefore got tan(arctan(p))=tan(a), and so p=tan(a)...I did the same for arctan(1/p) but with b, giving me (1/p)=tan(b)...
    I realize that the answer to this question should be 1 but have no idea how to get it. Any help would be greatly appreciated! Thanks!
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  2. #2
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    Hello, damanwithdaplan!

    If we look at it the right way, we don't need fancy formulas.


    Determine: . \sin\left[\arctan p+\arctan\frac{1}{p}\right]\;\text{ if }p>0

    Let: . \begin{Bmatrix} \alpha &=& \arctan p \\ \beta &=& \arctan\frac{1}{p} \end{Bmatrix} .[1]


    The problem becomes: . \sin(\alpha + \beta)


    From [1], we have: . \begin{Bmatrix}\tan\alpha &=& p \\ \tan\beta &=&\frac{1}{p}\end{Bmatrix}

    And we have this right triangle:
    Code:
                            *
                         * β|
                      *     |
                   *        | p
                *           |
             * α            |
          * - - - - - - - - *
                   1

    Get it?
    The two angles are from the same right triangle!
    . . That is, \alpha + \beta \:=\:90^o


    Therefore: . \sin(\alpha + \beta) \;=\;\sin(90^o) \;=\;1

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