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Thread: A few HW questions I dont get

  1. #1
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    A few HW questions I dont get

    Using the conditions below
    sinA=3/4, A lies in quadrant 2
    CosB=12/13, B lies in quadrant 1

    Find the following:

    Cos (A+B) Tan(A-B)

    Cos B/2



    Solve the equation in th interval [0,2pie)

    2cos^2x-3cosx+1




    Use Half angle forumlas to find the exact value of the expression

    cos 22.5


    cos^2 75 - sin^2 75
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by slaying_ice View Post
    Using the conditions below
    sinA=3/4, A lies in quadrant 2
    CosB=12/13, B lies in quadrant 1

    Find the following:

    Cos (A+B) Tan(A-B)

    Cos B/2
    First:
    $\displaystyle cos(A + B) = cos(A) cos(B) - sin(A) sin(B)$

    Second:
    $\displaystyle tan(A - B) = \frac{sin(A - B)}{cos(A - B)}$

    $\displaystyle tan(A - B) = \frac{sin(A)cos(B) - sin(B)cos(A)}{cos(A)cos(B) + sin(A)sin(B)}$

    Now:
    $\displaystyle sin(A) = \frac{3}{4}$ and is in the 2nd quadrant.

    Thus: $\displaystyle cos(A) = -\sqrt{1 - sin^2(A)} = -\frac{1}{2}$

    Again:
    $\displaystyle cos(B) = \frac{12}{13}$ and is in the first quadrant.

    Thus: $\displaystyle sin(B) = +\sqrt{1 - cos^2(B)} = \frac{5}{13}$

    To the problems:
    $\displaystyle cos(A + B) = cos(A) cos(B) - sin(A) sin(B)$

    $\displaystyle cos(A + B) = -\frac{1}{2} \cdot \frac{12}{13} - \frac{3}{4} \cdot \frac{5}{13} = -\frac{3}{4}$

    $\displaystyle tan(A - B) = \frac{sin(A)cos(B) - sin(B)cos(A)}{cos(A)cos(B) + sin(A)sin(B)}$

    $\displaystyle tan(A - B) = \frac{\frac{3}{4} \cdot \frac{12}{13} - \frac{5}{13} \cdot -\frac{1}{2}}{-\frac{1}{2} \cdot \frac{12}{13} + \frac{3}{4} \cdot \frac{5}{13}} = - \frac{46}{9}$

    So
    $\displaystyle cos(A + B) \cdot tan(A - B) = -\frac{3}{4} \cdot -\frac{46}{9} = \frac{23}{6}$
    ================================================== =======
    $\displaystyle cos \left ( \frac{B}{2} \right ) = \pm \sqrt{\frac{1 + cos(B)}{2}}$

    $\displaystyle cos \left ( \frac{B}{2} \right ) = \pm \sqrt{\frac{1 + \frac{12}{13}}{2}} = \pm \sqrt{\frac{25}{26}} = \pm \frac{5}{\sqrt{26}}$

    Since B is in the 1st quadrant, so is B/2. Thus
    $\displaystyle cos \left ( \frac{B}{2} \right ) = \frac{5 \sqrt{26}}{26}$

    -Dan
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by slaying_ice View Post
    Solve the equation in th interval [0,2pie)

    2cos^2x-3cosx+1
    By the way, the Greek letter $\displaystyle \pi$ is spelled "pi" not "pie."

    There's no equation to solve here!

    I'll assume you meant:
    $\displaystyle 2cos^2(x) - 3cos(x) + 1 = 0$

    Let's simplify a bit: Let $\displaystyle y = cos(x)$.
    The we have:
    $\displaystyle 2y^2 - 3y + 1 = 0$

    This factors:
    $\displaystyle (2y - 1)(y - 1) = 0$

    So $\displaystyle y = \frac{1}{2}$ or $\displaystyle y = 1$.

    Thus
    $\displaystyle cos(x) = \frac{1}{2}$ ==> $\displaystyle x = \frac{\pi}{3}, \frac{5 \pi}{3}$ rad

    or

    $\displaystyle cos(x) = 1$ ==> $\displaystyle x = 0$.

    So the full solution set on $\displaystyle [0, 2 \pi) $ is $\displaystyle x = 0, \frac{\pi}{3}, \frac{5 \pi}{3}$.

    -Dan
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    On the second problem you assumed correct on what the equation was.
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  5. #5
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by slaying_ice View Post
    Use Half angle forumlas to find the exact value of the expression

    cos 22.5


    cos^2 75 - sin^2 75
    The first one:
    $\displaystyle cos(22.5) = cos \left ( \frac{45}{2} \right ) = \sqrt{ \frac{1 + cos(45)}{2} } = \sqrt{ \frac{1 + \frac{ \sqrt{2} }{2} }{2} } = \frac{ \sqrt{2 + \sqrt{2} }}{2}$

    The second one actually isn't a half-angle formula, it's a double angle formula:
    $\displaystyle cos(150) = cos(2 \cdot 75) = cos^2(75) - sin^2(75)$

    and
    $\displaystyle cos(150) = -cos(30) = -\frac{\sqrt{3}}{2}$

    -Dan
    Last edited by topsquark; Dec 3rd 2006 at 04:33 PM.
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  6. #6
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    The file you uploaded to solve the double angle problem isnt showing up. Plus I have a couple more problems that need to be solved if you dont mind


    Find the exact valuse of Sin2x Cos2x and tan 2x given tan=4/3 in the first quadrant.



    and


    Find the exact valuse of sin x/2 cos x/2 and tan x/2 when cosx= -5/13 and x is in the third quadrant


    Thank you

    ~Kyle
    Last edited by slaying_ice; Dec 3rd 2006 at 04:35 PM.
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  7. #7
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by slaying_ice View Post
    The file you uploaded to solve the double angle problem isnt showing up.
    I fixed it. Sorry about that.

    -Dan
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    Plus I have a couple more problems that need to be solved if you dont mind


    Find the exact valuse of Sin2x Cos2x and tan 2x given tan=4/3 in the first quadrant.



    and


    Find the exact valuse of sin x/2 cos x/2 and tan x/2 when cosx= -5/13 and x is in the third quadrant


    Thank you

    ~Kyle
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  9. #9
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by slaying_ice View Post
    Plus I have a couple more problems that need to be solved if you dont mind


    Find the exact valuse of Sin2x Cos2x and tan 2x given tan=4/3 in the first quadrant.
    $\displaystyle tan(x) = \frac{4}{3}$ so

    $\displaystyle sin(x) = sin \left ( tan^{-1} \left ( \frac{4}{3} \right ) \right )$

    This is a 3 - 4 - 5 triangle, so
    $\displaystyle sin(x) = \frac{4}{5}$.

    Similarly:
    $\displaystyle cos(x) = \frac{3}{5}$

    So:
    $\displaystyle sin(2x) = 2 sin(x) cos(x) = 2 \cdot \frac{4}{5} \cdot \frac{3}{5} = \frac{24}{25}$

    $\displaystyle cos(2x) = cos^2(x) - sin^2(x) = \left (\frac{4}{5} \right ) ^2 - \left ( \frac{3}{5} \right ) ^2 = \frac{7}{25}$

    $\displaystyle tan(2x) = \frac{sin(2x)}{cos(2x)} = \frac{ \frac{24}{25} }{ \frac{7}{25} } = \frac{24}{7}$

    -Dan
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  10. #10
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by slaying_ice View Post
    Find the exact valuse of sin x/2 cos x/2 and tan x/2 when cosx= -5/13 and x is in the third quadrant
    $\displaystyle cos(x) = -\frac{5}{13}$

    Thus
    $\displaystyle sin \left ( \frac{x}{2} \right ) = \pm \sqrt{\frac{1 - cos(x)}{2}} = \pm \sqrt{ \frac{1 + \frac{5}{13}}{2} } = \pm \frac{3}{\sqrt{13}}$

    Since x is 3rd quadrant, x/2 will be 2nd quadrant.
    Thus $\displaystyle sin \left ( \frac{x}{2} \right ) = \frac{3 \sqrt{13}}{13}$

    $\displaystyle cos \left ( \frac{x}{2} \right ) = \pm \sqrt{\frac{1 + cos(x)}{2}} = \pm \sqrt{ \frac{1 - \frac{5}{13}}{2} } = \pm \frac{2}{\sqrt{13}}$

    Again, since x/2 is 2nd quadrant:
    $\displaystyle cos \left ( \frac{x}{2} \right ) = - \frac{2\sqrt{13}}{13}$

    $\displaystyle tan \left ( \frac{x}{2} \right ) = \frac{sin \left ( \frac{x}{2} \right )}{cos \left ( \frac{x}{2} \right )} = \frac{ \frac{3 \sqrt{13}}{13} }{ - \frac{2\sqrt{13}}{13} } = - \frac{3}{2}$

    -Dan
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    Thank you very much you have saved me from homework doom
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  12. #12
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by slaying_ice View Post
    Thank you very much you have saved me from homework doom
    But I LIKE Doom! (Doom, DoomII, etc...)

    -Dan
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