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Thread: Inverse Trigonometric Functions

  1. #1
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    Inverse Trigonometric Functions

    Establishing an Identity Involving Inverse Trigonometric Functions.

    (1) Show that tan[arcsin(v)] = v/(sqrt{1 - v^2})

    (2) Show that sin[arccos(v)] = sqrt{1 - v^2}
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  2. #2
    o_O
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    #1: Recall that $\displaystyle \tan \theta = \frac{\sin \theta}{\cos \theta}$

    So: $\displaystyle \tan \left(\arcsin v\right) = \frac{\sin \left(\arcsin v\right)}{\cos \left(\arcsin v\right)} = \frac{v}{\sqrt{1 - \big[\sin \left(\arcsin v\right)\big]^2}} = ...$


    #2: Similar to what was done in the denominator of #1, recall that $\displaystyle \cos^2 x = 1 - \sin^2x$

    So: $\displaystyle \sin \left(\arccos v\right) = \sqrt{1 - \big[\cos \left(\arccos v\right)\big]^2}$
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  3. #3
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by magentarita View Post
    Establishing an Identity Involving Inverse Trigonometric Functions.

    (1) Show that tan[arcsin(v)] = v/(sqrt{1 - v^2})

    (2) Show that sin[arccos(v)] = sqrt{1 - v^2}
    #1: If we were to plot a triangle that would represent $\displaystyle \sin^{-1}v$, the leg opposite the angle would have a value of v, and the hypotenuse would have a value of one.

    To solve for the side adjacent to the angle, apply pythagoras' theorem:

    $\displaystyle v^2+b^2=1\implies b=\sqrt{1-v^2}$

    Now, since $\displaystyle \tan\vartheta=\frac{opp}{adj}$, and $\displaystyle opp=v~and~adj=\sqrt{1-v^2}$ we see then that $\displaystyle \tan\vartheta=\tan\left(\sin^{-1}v\right)=\frac{v}{\sqrt{1-v^2}}$

    $\displaystyle \mathbb{Q.E.D.}$


    #2: If we were to plot a triangle that would represent $\displaystyle \cos^{-1}v$, the leg adjacent to the angle would have a value of v, and the hypotenuse would have a value of one.

    To solve for the side opposite to the angle, apply pythagoras' theorem:

    $\displaystyle v^2+b^2=1\implies b=\sqrt{1-v^2}$

    Now, since $\displaystyle \sin\vartheta=\frac{opp}{hyp}$, and $\displaystyle opp=\sqrt{1-v^2}~and~hyp=1$ we see then that $\displaystyle \sin\vartheta=\sin\left(\cos^{-1}v\right)=\sqrt{1-v^2}$

    $\displaystyle \mathbb{Q.E.D.}$

    Does this make sense??

    --Chris
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  4. #4
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    Thanks

    Thank you both. Of course, Chris' method is much easier to follow.
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    Hello, magentarita!

    Another approach . . .


    Establishing an identity involving inverse trigonometric functions.

    (1) Show that: .$\displaystyle \tan[\arcsin(v)] \:= \:\frac{v}{\sqrt{1 - v^2}}$
    Recall that an inverse trig expression is an angle.


    We have: .$\displaystyle \tan\underbrace{[\arcsin(v)]}_{\text{some angle }\theta}$

    That is: .$\displaystyle \theta$ is an angle whose sine is $\displaystyle v.$

    We have: .$\displaystyle \sin\theta \:=\:\frac{v}{1} \:=\:\frac{opp}{hyp}$

    $\displaystyle \theta$ is in a right triangle with: .$\displaystyle opp = v,\;hyp = 1$
    . . and we want: .$\displaystyle \tan\theta$
    Code:
                          *
                 1    *   |
                  *       | v
              * θ         |
          * - - - - - - - *
                 ____
                √1-vē
    Using Pythagorus, we find that .$\displaystyle adj = \sqrt{1-v^2}$


    Therefore: .$\displaystyle \tan\theta \:=\:\frac{opp}{adj} \:=\:\frac{v}{\sqrt{1-v^2}} $



    Ahh, Chris beat me to it . . .
    .
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  6. #6
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    soroban

    Quote Originally Posted by Soroban View Post
    Hello, magentarita!

    Another approach . . .

    Recall that an inverse trig expression is an angle.


    We have: .$\displaystyle \tan\underbrace{[\arcsin(v)]}_{\text{some angle }\theta}$

    That is: .$\displaystyle \theta$ is an angle whose sine is $\displaystyle v.$

    We have: .$\displaystyle \sin\theta \:=\:\frac{v}{1} \:=\:\frac{opp}{hyp}$

    $\displaystyle \theta$ is in a right triangle with: .$\displaystyle opp = v,\;hyp = 1$
    . . and we want: .$\displaystyle \tan\theta$
    Code:
                          *
                 1    *   |
                  *       | v
              * θ         |
          * - - - - - - - *
                 ____
                √1-vē
    Using Pythagorus, we find that .$\displaystyle adj = \sqrt{1-v^2}$


    Therefore: .$\displaystyle \tan\theta \:=\:\frac{opp}{adj} \:=\:\frac{v}{\sqrt{1-v^2}} $


    Ahh, Chris beat me to it . . .
    .
    Thank you Soroban.
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