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Thread: Trig Inverse and Differentiation

  1. #1
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    Trig Inverse and Differentiation

    Prove $\displaystyle cos^{-1} 2x + sin^{-1} 2x = a $


    $\displaystyle \frac{-1}{2}\leq x \leq \frac{1}{2}$

    thank you.
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  2. #2
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    Re: Trig Inverse and Differentiation

    $\displaystyle \begin{aligned}\bigg[\bigg(\sin^{-1}{2x}\bigg)+\bigg(\cos^{-1}{2x}\bigg)\bigg]' & = \bigg(\sin^{-1}{2x}\bigg)'+\bigg(\cos^{-1}{2x}\bigg)' \\& = \frac{(2x)'}{\sqrt{1-(2x)^2}}-\frac{(2x)'}{\sqrt{1-(2x)^2}} \\& = 0.\end{aligned}$

    Alternatively, you can find the exact value of $\displaystyle a$ by usual trigonometry.
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  3. #3
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    Re: Trig Inverse and Differentiation

    how do i find the value of a?
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  4. #4
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    Re: Trig Inverse and Differentiation

    Hello, BabyMilo!

    $\displaystyle \text{Prove: }\:\cos^{-1}(2x) + \sin^{-1}(2x) \:=\: a \qquad \text{-}\tfrac{1}{2}\:\leq\: x\: \leq\: \tfrac{1}{2}$

    Let $\displaystyle \alpha \:=\:\cos^{-1}(2x) \quad\Rightarrow\quad \cos\alpha \,=\, 2x\;\;[1] $

    Let $\displaystyle \beta \:=\:\sin^{-1}(2x) \quad\Rightarrow\quad \sin\beta \,=\,2x \;\;[2] $


    From [1]: .$\displaystyle \cos\alpha \:=\:\frac{2x}{1} \:=\:\frac{adj}{hyp}$

    Hence, $\displaystyle \alpha$ is in this right triangle.

    Code:
                      *
               1   *  *
                *     *
             * α      *
          *  *  *  *  *
                2x

    From [2]: .$\displaystyle \sin\beta \:=\:\frac{2x}{1} \:=\:\frac{opp}{hyp}$

    Hence, $\displaystyle \beta$ is in this right triangle:

    Code:
                *
               **
           1  * *
             *  * 2x
            *   *
           * β  *
          * * * *

    We see that the two triangles are congruent.

    That is, $\displaystyle \alpha$ and $\displaystyle \beta$ are in the same right triangle!

    Code:
                      *
               1   * β*
                *     *
             * α      *
          *  *  *  *  *
                2x

    Therefore: .$\displaystyle \alpha + \beta \:=\:\frac{\pi}{2} \quad\Rightarrow\quad \cos^{-1}(2x) + \sin^{-1}(2x) \:=\:\frac{\pi}{2}$

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