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Math Help - changing the term

  1. #1
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    changing the term

    x=3sinu

    How can I write 1/3 sec u in terms of x?
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  2. #2
    Junior Member Youkla's Avatar
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    Re: changing the term

    So if I am understanding correctly, you want to write

    \frac{1}{3}sec(u) in terms of x where you use the equation x=3sin(u)

    Well, we know that \sec(u) = \frac{1}{\cos(u)}, thus \frac{1}{3}sec(u) = \frac{1}{\3cos(u)}

    We also know the mother of all identities is \sin^2(u) + \cos^2(u) = 1 which can be solved for \cos(u) as \cos(u) = \pm\sqrt{1-\sin^2(u)}

    Now, using the equation x=3sin(u) do you think you can finish this?
    Last edited by Youkla; October 20th 2011 at 11:21 AM.
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  3. #3
    Member anonimnystefy's Avatar
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    Re: changing the term

    hi Youkla

    i think it should be \pm\sqrt{1-\sin^2{x}}
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  4. #4
    Junior Member Youkla's Avatar
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    Re: changing the term

    Quote Originally Posted by anonimnystefy View Post
    hi Youkla

    i think it should be \pm\sqrt{1-\sin^2{x}}
    In the words of Storage Wars..."Yuuuuuuuup!"

    Thanks
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  5. #5
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    Re: changing the term

    Hello, Stuck Man!

    \text{Given: }\:x \:=\:3\sin u

    \text{Write }\tfrac{1}{3}\sec u\text{ in terms of }x.

    We are given: . x \:=\:3\sin u \quad\Rightarrow\quad \sin u \:=\:\frac{x}{3} \:=\:\frac{opp}{hyp}

    That is, u is an angle in a right triangle with: opp = x,\;hyp = 3
    Using Pythagorus, we find that: . adj = \sqrt{9-x^2}

    Hence: . \sec u \:=\:\frac{hyp}{adj} \:=\:\frac{3}{\sqrt{9-x^2}}

    Therefore: . \frac{1}{3}\sec u \;=\;\frac{1}{3}\cdot\frac{3}{\sqrt{9-x^2}} \;=\; \frac{1}{\sqrt{9-x^2}}

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  6. #6
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    Re: changing the term

    I don't think the plus/minus sign is justified. A root of that expression can't be negative. Thanks for the help. Sometimes I have wondered why people drop the plus/minus sign.
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  7. #7
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    Re: changing the term

    Quote Originally Posted by Stuck Man View Post
    I don't think the plus/minus sign is justified. A root of that expression can't be negative. Thanks for the help. Sometimes I have wondered why people drop the plus/minus sign.
    Actually, the " " is justified. The cosine function (and thus the secant function) can have the same or opposite sign as the sine function, depending upon the quadrant.

    Also, when Soroban found that adj = \sqrt{9-x^2}, he found it by taking the square root of (adj)^2, so it should have been adj = \pm\sqrt{9-x^2}, if you prefer to look at it this way.
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