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Math Help - Problem getting a Trig question check in My Calc

  1. #1
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    Problem getting a Trig question check in My Calc

    For example if I am given:

    \text{G:}Cos(x)=\frac{-1}{6} and x is in \text{Quadrant III}

    \text{U:}Sin\frac{x}{2}=? in quadrant III

    I worked it out in by using the half angle formula and got \frac{-\sqrt 21}{6}

    I usually use the old inverse trick on the calculator and check to see if the decimal I get matches the answer I got. But on several half angle problems I am getting an undefined message on my calc. What is the deal and how can I go about checking these...

    Like what PerfectHacker did on this post http://www.mathhelpforum.com/math-he...i-89-calc.html

    ----------
    Also one other question.... When I am using the double angle and half angle formulas, when I get what the single trig value is equal to such as Cos(x)=\frac{-1}{6} and I am going to stick the value in a problem do I want to insert the negative or positive sign along with it depending on which quadrant I am talking about? For example in the problem above we are in the 3rd quadrant so both Sin & Cos both are negative. Would I want to insert them into the formulas as negative? But suppose we are in quadrant 1 where both Sin & Cos are positive. Would I want to put both values into the double or half-angle formulas as a postive?


    I could use some real help with both questions if you could...


    Thanks as always for the help!
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  2. #2
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    Hello, qbkr21!

    Be careful! . . .


    Given: . \cos x =-\frac{1}{6} and x is in Q3

    Find: . \sin\frac{x}{2}

    Identity: . \sin\frac{\theta}{2}\:=\:\pm\sqrt{\frac{1-\cos\theta}{2}}

    So we have: . \sin\frac{x}{2} \:=\:\pm\sqrt{\frac{1 - (-\frac{1}{6})}{2}} \:=\:\pm\frac{\sqrt{21}}{6}
    . . . which you already knew


    Since x is in Q3\!:\;\;\pi \:\leq \:x \:\leq \:\frac{3\pi}{2}\quad\Rightarrow\quad \text{then: }\frac{\pi}{2} \:\leq \:\frac{x}{2} \:\leq \:\frac{3\pi}{4}

    . . Hence, \frac{x}{2} is in Quadrant 2, where sine is positive.

    Therefore: . \sin\frac{x}{2}\:= + \frac{\sqrt{21}}{6}

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  3. #3
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    Re:

    Thanks Soroban,


    So can I check all questions like these in my calculator after doing the work to ensure that my answers are correct. If the decimals matches the irrational fraction I got then I guess that my answers are correct?
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