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Thread: I need help asap i would appreciate it if anybody could help me thanks!!!!

  1. March 6th 2009, 06:30 PM #1
    angelica2619
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    Smile I need help asap i would appreciate it if anybody could help me thanks!!!!

    I need help with these problems....

    1. Is cosX=1/4 and 0<x<90, find cscx

    2. If cotx=square root of -6/3, and 90<x<180, find tan

    3.Simplify tanx/sinx

    4.simplify cotxtanx+sinxsecx
    Last edited by mr fantastic; April 25th 2009 at 05:23 AM. Reason: Distributed remaining questions to new threads
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  2. March 6th 2009, 06:47 PM #2
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    Quote Originally Posted by angelica2619 View Post
    I need help with these problems....

    1. Is cosX=1/4 and 0<x<90, find cscx

    [snip]
    1. \csc{x} = \frac{1}{\sin{x}}

    Remember the Pythagorean Identity...

    \cos^2{x} + \sin^2{x} = 1

    And since \cos{x} = \frac{1}{4}

    \left(\frac{1}{4}\right)^2 + \sin^2{x} = 1

    \frac{1}{16} + \sin^2{x} = 1

    \sin^2{x} = \frac{15}{16}

    \sin{x} = \sqrt{\frac{15}{16}}

    \sin{x} = \frac{\sqrt{15}}{4}

    \frac{1}{\sin{x}} = \frac{4}{\sqrt{15}}

    \frac{1}{\sin{x}} = \frac{4\sqrt{15}}{15}

    \csc{x} = \frac{4\sqrt{15}}{15}.

    Quote Originally Posted by angelica2619 View Post
    [snip]
    2. If cotx=square root of -6/3, and 90<x<180, find tan

    [snip]

    2. You can't have the square root of a negative number.
    Last edited by mr fantastic; April 25th 2009 at 05:25 AM.
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  3. March 6th 2009, 06:51 PM #3
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    Quote Originally Posted by angelica2619 View Post
    [snip]
    3.Simplify tanx/sinx

    [snip]
    3. Remember that \tan{x} = \frac{\sin{x}}{\cos{x}}.

    So \tan{x} \div \sin{x} = \frac{\sin{x}}{\cos{x}} \div \sin{x}

    = \frac{\sin{x}}{\cos{x}} \times \frac{1}{\sin{x}}

    = \frac{\sin{x}}{\sin{x}\cos{x}}

    = \frac{1}{\cos{x}}

    = \sec{x}.
    Last edited by mr fantastic; April 25th 2009 at 05:25 AM.
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  4. March 6th 2009, 06:54 PM #4
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    Quote Originally Posted by angelica2619 View Post
    [snip]
    4.simplify cotxtanx+sinxsecx

    [snip]
    4. Remember that \cot{x} = \frac{1}{\tan{x}} and \sec{x} = \frac{1}{\cos{x}}.

    So \cot{x}\tan{x} = \frac{1}{\tan{x}}\tan{x} = 1

    and \sin{x}\sec{x} = \sin{x}\frac{1}{\cos{x}} = \tan{x}.

    Therefore \cot{x}\tan{x} + \sin{x}\sec{x} = 1 + \tan{x}.
    Last edited by mr fantastic; April 25th 2009 at 05:26 AM.
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