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Math Help - Finding x in Trig

  1. #1
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    Finding x in Trig

    Question:
    Find sin(arccos(x))/cos(arcsin(x))



    I am not sure if this helps to solve anything, and im not sure if this si correct or not.
    =tan(arccos(x)/arcsin(x)
    =tan(arccot(x))
    =x
    Thank you
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  2. #2
    MHF Contributor MarkFL's Avatar
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    Re: Finding x in Trig

    Hint: use the identity \arcsin(x)+\arccos(x)=\frac{\pi}{2}...
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  3. #3
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    Re: Finding x in Trig

    Hello, DylanRenke!

    \text{Find: }\:\frac{\sin(\arccos x)}{\cos(\arcsin x )}

    Let: . \begin{Bmatrix}\alpha &=& \arccos x & [1] \\ \beta &=& \arcsin x & [2] \end{Bmatrix}

    The problem becomes: . \frac{\sin\alpha}{\cos\beta} .(a)


    From [1]: / \alpha \:=\:\arccos x \quad\Rightarrow\quad \cos\alpha \,=\,x

    We have: . \cos\alpha \,=\,\frac{x}{1} \,=\,\frac{adj}{hyp}

    We have a right triangle with: adj = x,\:hyp = 1.

    Pythagorus tells us that: . opp \,=\,\sqrt{1-x^2}

    Hence: . \sin\alpha \:=\:\frac{opp}{hyp} \:=\:\frac{\sqrt{1-x^2}}{1} \:=\:\sqrt{1-x^2} .(b)


    From [2]: . \beta \,=\,\arcsin x \quad\Rightarrow\quad \sin\beta \,=\,x

    We have: . \sin\beta \,=\,\frac{x}{1} \,=\,\frac{opp}{hyp}

    We have a right triangle with: opp = x.\:hyp \,=\,1.

    Pythagorus tells us that: . adj \,=\,\sqrt{1-x^2}

    Hence: . \cos\beta \:=\:\frac{adj}{hyp} \:=\:\frac{\sqrt{1-x^2}}{1} \:=\:\sqrt{1-x^2}} .(c)


    Substitute (b) and (c) into (a): . \frac{\sin\alpha}{\cos\beta} \;=\;\frac{\sqrt{1-x^2}}{\sqrt{1-x^2}} \;=\;1
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  4. #4
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    Re: Finding x in Trig

    Quote Originally Posted by DylanRenke View Post
    Question:
    Find sin(arccos(x))/cos(arcsin(x))



    I am not sure if this helps to solve anything, and im not sure if this si correct or not.
    =tan(arccos(x)/arcsin(x)
    =tan(arccot(x))
    This is wrong. "arccos(x)/arcsin(x)" is NOT "arccot(x)". Just because a function has some property doesn't mean its inverse has that property!

    =x
    Thank you
    Notice that sin(u)= \pm\sqrt{1- cos^2(u)} and so sin(arccos(x))= \pm\sqrt{1- cos^2(arccos(x)}= \pm\sqrt{1- x^2}
    However, cosine and sine are not "one to one" functions so they do NOT, strictly speaking, have inverses. In order to be able to talk about "arccos(x)" and "arcsin(x)" we hage to restrict cosine to 0 to \pi and sine to -\pi/2 to [tex]\pi/2[/itex]. With those restrictions, sine and cosine are both postive and so both cos(arcsin(x)) and sin(arccos(x)) are equal to [tex]\sqrt{1- x^2}[tex].
    Last edited by HallsofIvy; October 14th 2012 at 12:40 PM.
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  5. #5
    MHF Contributor MarkFL's Avatar
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    Re: Finding x in Trig

    If you use the hint I gave, you find:

    \frac{ \sin( \arccos(x))}{ \cos( \arcsin(x))}= \frac{ \sin( \arccos(x))}{ \cos\left( \frac{\pi}{2}-\arccos(x) \right)}= \frac{ \sin( \arccos(x))}{ \sin( \arccos(x))}=1
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  6. #6
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    Re: Finding x in Trig

    Thank you everybody.
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