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Math Help - Inverse trigonometry

  1. #1
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    Inverse trigonometry

    {{tan}^{-1}} \frac{a}{x}  +  {{tan}^{-1}} \frac{b}{x} = \frac{x}{2}


    Find the value of x.
    This doesn't seem to be striking me at all. I used the trigonometry identity of tan inverse A + tan inverse B, but the answer doesn't really spring from that..
    Any help will be appreciated.

    Also,
    tan({cos}^{-1}\frac{1}{5{\sqrt{2}}} - {sin}^{-1}\frac{4}{\sqrt{17}})

    Find the value of the expression here. I tried substituting and converting cos inverse to sin inverse but the identity is becoming too long and the answer ambiguous.
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  2. #2
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    Quote Originally Posted by SVXX View Post
    {{tan}^{-1}} \frac{a}{x}  +  {{tan}^{-1}} \frac{b}{x} = \frac{x}{2}
    Well, [tex]\theta= tan^{-1} \frac{b}{x}[\math] if and only if \theta is an angle in a right triangle with "opposite side" of length a and "near side" of length x. Similarly, if \phi= tan^{-1}\frac{b}{x} if and only if [itex]\phi[/itex] is an angle in a right triangle with "opposite side" b and "near side" x. Now, imagine those two triangles sharing the side of length x. You have a triangle with angle \theta+ \phi= tan^{-1}\frac{a}{x}+ \frac{b}{x}= \frac{x}{2}, base of lenght a+b and altitude of length x. Unfortunately, I don't really see where to go from there.


    Find the value of x.
    This doesn't seem to be striking me at all. I used the trigonometry identity of tan inverse A + tan inverse B, but the answer doesn't really spring from that..
    Any help will be appreciated.

    Also,
    tan({cos}^{-1}\frac{1}{5{\sqrt{2}}} - {sin}^{-1}\frac{4}{\sqrt{17}})

    Find the value of the expression here. I tried substituting and converting cos inverse to sin inverse but the identity is becoming too long and the answer ambiguous.[/QUOTE]
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  3. #3
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    Hello, SVXX!

    Here's the second one . . .


    Simplify: . \tan\left[\cos^{-1}\!\left(\tfrac{1}{5{\sqrt{2}}}\right) - \sin^{-1}\!\left(\tfrac{4}{\sqrt{17}}\right)\right]
    \text{We have: }\;\tan\left[\underbrace{\cos^{-1}\!\left(\tfrac{1}{5\sqrt{2}}\right)}_{\alpha} - \underbrace{\sin^{-1}\!\left(\tfrac{4}{\sqrt{17}}\right)}_{\beta}\rig  ht] .[1]


    \alpha \:=\:\cos^{-1}\left(\tfrac{1}{5\sqrt{2}}\right) \quad\Rightarrow\quad \cos\alpha \:=\:\tfrac{1}{5\sqrt{2}} \:=\:\frac{adj}{hyp} \quad\Rightarrow\quad opp \:=\:7
    . . \tan\alpha \:=\:\frac{opp}{adj} \:=\:\tfrac{7}{1} \:=\:7 \quad\Rightarrow\quad \boxed{\alpha \:=\:\tan^{-1}(7)}

    \beta \:=\:\sin^{-1}\left(\frac{4}{\sqrt{17}}\right) \quad\Rightarrow\quad \sin\beta \:=\:\frac{4}{\sqrt{17}} \:=\:\frac{opp}{hyp} \quad\Rightarrow\quad adj \:=\:1
    . . \tan\beta \:=\:\frac{opp}{adj} \:=\:\frac{4}{1} \:=\:4 \quad\Rightarrow\quad \boxed{\beta \:=\:\tan^{-1}(4)}


    Substitute into [1]:

    \tan[\alpha - \beta] \;=\;\tan\left[\tan^{-1}(7) - \tan^{-1}(4)\right]

    . . . . . . . = \;\frac{\tan[\tan^{-1}(7)] - \tan[\tan^{-1}(4)]}{1 - [\tan^{-1}(7)][\tan^{-1}(4)]}

    . . . . . . . = \;\frac{7-4}{1-7\!\cdot\!4} \;=\;\frac{\text{-}3}{\text{-}27} \;=\;\boxed{\frac{1}{9}}

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  4. #4
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    Ah mighty thanks Soroban! So the trick was to directly convert both into inverse tan.
    What about the first?

    EDIT : The scanned question paper was just too dark to see, I'm sorry. I made a mistake in posting the first question, it's pi/2 not x/2. That makes it easier.
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