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Thread: 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?

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

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

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

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


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

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


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

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


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

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


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

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


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

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


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

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


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

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


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

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


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

    thanks
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