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Thread: How do you simplify this problem?

  1. #1
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    How do you simplify this problem?

    $\displaystyle sin(\frac{-3pi}{4}) cos(\frac{5pi}{6})$



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  2. #2
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    Quote Originally Posted by Darkhrse99 View Post
    $\displaystyle sin(\frac{-3pi}{4}) cos(\frac{5pi}{6})$
    Didn't I tell you before to memorize certain trigonometric values? It seems you have not fulfilled this assignment.

    sin(-3pi/4) = -sin(3pi/4) = -sin(pi/4) = ???

    cos(5pi/6) = -cos(pi/6) = ???

    Now what?

    Note: If you're using LaTeX, feel free to put a slash in front of those "pi"s. Excellent results can be obtained.
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  3. #3
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    Quote Originally Posted by TKHunny View Post
    Didn't I tell you before to memorize certain trigonometric values? It seems you have not fulfilled this assignment.

    sin(-3pi/4) = -sin(3pi/4) = -sin(pi/4) = ???$\displaystyle \frac{\sqrt1}{2}$?

    cos(5pi/6) = -cos(pi/6) = ???$\displaystyle \frac{1}{2}$?

    Now what?

    Note: If you're using LaTeX, feel free to put a slash in front of those "pi"s. Excellent results can be obtained.
    Is this correct?
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  4. #4
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    Sorry, it's a little like you are just throwing darts.

    On the first:

    1) Where did the negative sign go?

    2) $\displaystyle \sqrt{1}$?

    On the second:

    3) Where did the negative sign go?

    4) Please note the difference between $\displaystyle \frac{\pi}{6}$ and $\displaystyle \frac{\pi}{3}$.

    Look at a chart of values. Think it through. Don't just stab at it.
    Last edited by TKHunny; Jun 19th 2008 at 04:51 AM.
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  5. #5
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    So the first on is$\displaystyle \frac{-\sqrt1}{2}$ and the second one is. $\displaystyle \frac{-\sqrt3} {2}$. I'm finding the (-)(Y value for $\displaystyle \frac{\pi}{4}$
    and the X value for $\displaystyle \frac{\pi}{4}$ Is this correct. I'm not trying to take stabs at it, but i truely don't know how to do the problem.
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  6. #6
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    Is this answer correct?
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  7. #7
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    You must MEMORIZE a few important values.

    $\displaystyle 0\;\;\sin(0)\;=\;0\;\;\cos(0)\;=\;1$

    $\displaystyle \frac{\pi}{6}\;\;\sin(\frac{\pi}{6})\;=\;\frac{1}{ 2}\;\;\cos(\frac{\pi}{6})\;=\;\frac{\sqrt{3}}{2}$

    $\displaystyle \frac{\pi}{4}\;\;\sin(\frac{\pi}{4})\;=\;\frac{\sq rt{2}}{2}\;\;\cos(\frac{\pi}{4})\;=\;\frac{\sqrt{2 }}{2}$

    $\displaystyle \frac{\pi}{3}\;\;\sin(\frac{\pi}{3})\;=\;\frac{\sq rt{3}}{2}\;\;\cos(\frac{\pi}{3})\;=\;\frac{1}{2}$

    $\displaystyle \frac{\pi}{2}\;\;\sin(\frac{\pi}{2})\;=\;1\;\;\cos (\frac{\pi}{2})\;=\;0$

    There is NO substitute for memorizing this tableau. Put it in your brain.

    Note: You still have $\displaystyle \sqrt{1}$. Don't write things that make no sense.
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  8. #8
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    Thanks for the help.

    My text book shows that$\displaystyle \frac{\sqrt1}{2}$ is the same as $\displaystyle \frac{\sqrt2}{2}$
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  9. #9
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    So really what I'm trying to do when simplifying is getting the x or y of the above table?


    So in the problem below I need to find the X or Y value?
    Simplify $\displaystyle tan(\frac{-2\pi}{3})csc(\frac{3\pi}{4})$
    Tan=$\displaystyle \frac{Y}{X}$ CSC=$\displaystyle \frac{1}{Y}$ Do I find the x or y value for tan and cosecant?
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  10. #10
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    Quote Originally Posted by Darkhrse99 View Post
    Thanks for the help.

    My text book shows that$\displaystyle \frac{\sqrt1}{2}$ is the same as $\displaystyle \frac{\sqrt2}{2}$
    No, once again, it's not $\displaystyle \frac{\sqrt1}{2}$ but $\displaystyle \sqrt{\frac 12}$ which is the same as $\displaystyle \frac{\sqrt2}{2}$
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  11. #11
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    or maybe $\displaystyle \frac{1}{\sqrt{2}}$.

    Notation means stuff. Don't disregard it for expediency.
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  12. #12
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    Quote Originally Posted by Darkhrse99 View Post
    So really what I'm trying to do when simplifying is getting the x or y of the above table?


    So in the problem below I need to find the X or Y value?
    Simplify $\displaystyle tan(\frac{-2\pi}{3})csc(\frac{3\pi}{4})$
    Tan=$\displaystyle \frac{Y}{X}$ CSC=$\displaystyle \frac{1}{Y}$ Do I find the x or y value for tan and cosecant?
    What does your memorized table tell you?
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