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Math Help - trigo inverse

  1. #1
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    trigo inverse

    solve x 2tan(inverse)(sin x)=tan(inverse)(2sec x) 0<x<90
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  2. #2
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    Is this your equation?

    \displaystyle 2\tan^{-1}{(\sin{x})} = \tan^{-1}{(2\sec{x})} for \displaystyle 0^{\circ} < x < 90^{\circ}...
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  3. #3
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    Hello, prasum!

    \text{Solve for }x\!:\;2\tan^{\text{-}1}(\sin x) \;=\; \tan^{\text{-}1}(2\sec x),\;\text{ for }0^o < x < 90^o

    Let: . \begin{Bmatrix}\alpha &=& \tan^{\text{-}1}(\sin x)  & \Rightarrow& \tan\alpha &=& \sin x \\ \beta &=& \tan^{\text{-}1}(2\sec x) & \Rightarrow & \tan\beta &=& 2\sec x \end{Bmatrix} .[1]

    And the equation becomes: . 2\alpha \:=\:\beta


    Take the tangent of both sides: . \tan2\alpha \;=\;\tan\beta

    . . And we have: . \displaystyle \frac{2\tan\alpha}{1 - \tan^2\alpha} \;=\;\tan\beta


    Substitute [1]: . \dfrac{\sin x}{1-\sin^2\!x} \;=\;2\sec x

    . . And we have: . \dfrac{\sin x}{\cos^2\!x} \;=\;\dfrac{2}{\cos x}


    Since x \ne 90^o,\;\cos x \ne 0

    \text{Multiply by }\cos x\!:\;\;\dfrac{\sin x}{\cos x} \;=\;2 \quad\Rightarrow\quad \tan x \:=\:2


    Therefore: . x \;=\;\tan^{\text{-}1}2 \;\approx\;63.4^o

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  4. #4
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    Quote Originally Posted by Soroban View Post
    Hello, prasum!


    Let: . \begin{Bmatrix}\alpha &=& \tan^{\text{-}1}(\sin x)  & \Rightarrow& \tan\alpha &=& \sin x \\ \beta &=& \tan^{\text{-}1}(2\sec x) & \Rightarrow & \tan\beta &=& 2\sec x \end{Bmatrix} .[1]

    And the equation becomes: . 2\alpha \:=\:\beta


    Take the tangent of both sides: . \tan2\alpha \;=\;\tan\beta

    . . And we have: . \displaystyle \frac{2\tan\alpha}{1 - \tan^2\alpha} \;=\;\tan\beta


    Substitute [1]: . \dfrac{\sin x}{1-\sin^2\!x} \;=\;2\sec x

    . . And we have: . \dfrac{\sin x}{\cos^2\!x} \;=\;\dfrac{2}{\cos x}


    Since x \ne 90^o,\;\cos x \ne 0

    \text{Multiply by }\cos x\!:\;\;\dfrac{\sin x}{\cos x} \;=\;2 \quad\Rightarrow\quad \tan x \:=\:2


    Therefore: . x \;=\;\tan^{\text{-}1}2 \;\approx\;63.4^o

    thanks
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