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Math Help - Inverse trigonometric Function Question 2

  1. #1
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    Exclamation Inverse trigonometric Function Question 2

    Prove that
    arc sinx + arc cosx = arc tan [xy+{(1-x^2)(1-y^2)}^1/2]/[y{(1-x^2)}^1/2 -x{(1-y^2)}^1/2]
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  2. #2
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    Trig equation

    Hello tariq_h_tauheed
    Quote Originally Posted by tariq_h_tauheed View Post
    Prove that
    arc sinx + arc cosx = arc tan [xy+{(1-x^2)(1-y^2)}^1/2]/[y{(1-x^2)}^1/2 -x{(1-y^2)}^1/2]
    I'm sure you mean \arcsin x + \arccos y= ....

    So let a = \arcsin x, b = \arccos y.

    \Rightarrow x = \sin a,\quad y = \cos b

    \Rightarrow \cos a = (1-\sin^2a)^{\tfrac12} = (1-x^2)^{\tfrac12}, \quad\sin b = (1-y^2)^{\tfrac12}

    Then \tan(a+b) = \frac{\sin(a+b)}{\cos(a+b)}

    =\frac{\sin a\cos b+\cos a\sin b}{\cos a\cos b - \sin a \sin b}

    \Rightarrow \arcsin x + \arccos y = a + b = \arctan\Big(\frac{\sin a\cos b+\cos a\sin b}{\cos a\cos b - \sin a \sin b}\Big)

    =\arctan\Bigg(\frac{xy+(1-x^2)^{\tfrac12}(1-y^2)^{\tfrac12}}{y(1-x^2)^{\tfrac12}-x(1-y^2)^{\tfrac12}}\Bigg)

    Grandad
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  3. #3
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    Hello, tariq_h_tauheed!

    Another approach . . .


    Prove that: . \arcsin x + \arccos y \:=\: \arctan \left[\frac{xy+\sqrt{(1-x^2)(1-y^2)}} {y\sqrt{1-x^2} -x\sqrt{1-y^2}}\right]

    Let: . \begin{array}{ccccccc}\alpha \:=\:\arcsin x & \Rightarrow & \sin\alpha \:=\:x \\<br />
\beta \:=\:\arccos y & \Rightarrow & \cos\beta \:=\:y \end{array}

    Then we have: . \tan\alpha \;=\;\frac{x}{\sqrt{1-x^2}} \qquad \tan\beta \;=\; \frac{\sqrt{1-y^2}}{y} .[1]


    The left side is: . \alpha + \beta

    Take the tangent: . \tan(\alpha + \beta) \;=\;\frac{\tan\alpha + \tan\beta}{1 - \tan\alpha\tan\beta} .[2]


    Substitute [1] into [2]: . \tan(\alpha + \beta) \;=\;\frac{\dfrac{x}{\sqrt{1-x^2}} + \dfrac{\sqrt{1-y^2}}{y}} {1 - \dfrac{x}{\sqrt{1-x^2}}\cdot\dfrac{\sqrt{1-y^2}}{y}}


    Multiply top and bottom by \left(y\sqrt{1-x^2}\right)

    . . \tan(\alpha + \beta) \;=\;\frac{y\sqrt{1-x^2}\left(\frac{x}{\sqrt{1-x^2}} + \frac{\sqrt{1-y^2}}{y}\right)} {y\sqrt{1-x^2}\left(1 - \frac{x}{\sqrt{1-x^2}}\cdot\frac{\sqrt{1-y^2}}{y}\right)}\;= . \frac{xy + \sqrt{1-x^2}\sqrt{1-y^2}}{y\sqrt{1-x^2} - x\sqrt{1-y^2}}


    Therefore: . \alpha + \beta \;=\;\arcsin x + \arccos y \;=\;\arctan\left[\frac{xy + \sqrt{1-x^2}\sqrt{1-y^2}}{y\sqrt{1-x^2} - x\sqrt{1-y^2}}\right]

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  4. #4
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    Thanks a ton...
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