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Math Help - 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 \tan \theta = \frac{\sin \theta}{\cos \theta}

    So: \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 \cos^2 x = 1 - \sin^2x

    So: \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 \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:

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

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

    \mathbb{Q.E.D.}


    #2: If we were to plot a triangle that would represent \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:

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

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

    \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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  5. #5
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    Hello, magentarita!

    Another approach . . .


    Establishing an identity involving inverse trigonometric functions.

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


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

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

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

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


    Therefore: . \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: . \tan\underbrace{[\arcsin(v)]}_{\text{some angle }\theta}

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

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

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


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


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