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Thread: Trigo question. Help needed. Need to pass up on tomorrow.

  1. #1
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    Trigo question. Help needed. Need to pass up on tomorrow.

    Given that $\displaystyle \cos x = -\frac{2}{3}$ and $\displaystyle \sin y = -\frac{1}{\sqrt 6}$, that $\displaystyle 0^\circ\le x \le360^\circ$ and that $\displaystyle x$ and $\displaystyle y$ are in the same quadrant, find, without calculator, the values of $\displaystyle \cos\frac{x}{2}$.



    Man. This question is killing me. I only get to know $\displaystyle x$ and $\displaystyle y$ are in quadrant 3.
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  2. #2
    Super Member flyingsquirrel's Avatar
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    Hi
    Quote Originally Posted by simonsong View Post
    Given that $\displaystyle \cos x = -\frac{2}{3}$ and $\displaystyle \sin y = -\frac{1}{\sqrt 6}$, that $\displaystyle 0^\circ\le x \le360^\circ$ and that $\displaystyle x$ and $\displaystyle y$ are in the same quadrant, find, without calculator, the values of $\displaystyle \cos\frac{x}{2}$.


    Man. This question is killing me. I only get to know $\displaystyle x$ and $\displaystyle y$ are in quadrant 3.
    $\displaystyle \cos x=\cos \left(\frac{x}{2}+\frac{x}{2}\right)=2\cos^2\frac{ x}{2}-1$ hence $\displaystyle \cos \frac{x}{2}=\ldots$

    This should give you two values for $\displaystyle \cos \frac{x}{2}$. As you also know in which quadrant is $\displaystyle \frac{x}{2}$, you can choose the right solution among the two you've found.
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  3. #3
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    Omg. Never thought of that. Haha. Really thanks pal.
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