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Math Help - Some question that need help.

  1. #1
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    Some question that need help.

    1)Without evaluating the angles A and B,find the value of tan(A+2B),
    given tanA = 2/3 and tanB =1/3.

    2)Simplify (x/y)powerof2/3 . xpowerof3/ypowerof4

    3)Find all angles between 0deg and 360deg inclusive which satisfy the equation : 3cossquartA - 2cosA - 1 = 0

    4)find the value of x when : log(base2)x + log(base8)x = 4/3


    BIG thanks!!!
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  2. #2
    Member Glaysher's Avatar
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    1.

    Use tan addition formula

    \tan{(a+b)}=\frac{\tan{a}+\tan{b}}{1-\tan{a}\tan{b}}
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  3. #3
    Member Glaysher's Avatar
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    2.

    \left(\frac{x}{y}\right) ^{\frac{2}{3}} \frac{x^3}{y^4}

    \frac{x^{2/3}}{y^{2/3}}  \frac{x^3}{y^4}

    \left(\frac{x^{11}}{y^{14}}\right)^{\frac{1}{3}}
    Last edited by CaptainBlack; September 3rd 2006 at 04:41 AM. Reason: tidying up TeX
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  4. #4
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    Quote Originally Posted by Glaysher
    1.

    Use tan addition formula

    \tan{(a+b)}=\frac{\tan{a}+\tan{b}}{1-\tan{a}\tan{b}}
    You'd also need this trig identity seeing it is  tan(A+2B)

     tan 2B= \frac{2tanB}{1-\tan^2B}

    obviously you can sub that into the other identity given to give:

    \tan{(a+2b)}=\frac{\tan{a}+\frac{2tanB}{1-\tan^2B}}{1-\tan{a}\frac{2tanB}{1-\tan^2B}}
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  5. #5
    Member Glaysher's Avatar
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    4.

    \log_2{x} + \log_8{x} =\frac{4}{3}

    \log_2{x} + \frac{\log_2{x}}{\log_2{8}} =\frac{4}{3}

    \log_2{x} + \frac{\log_2{x}}{3} =\frac{4}{3}

    \frac{4}{3}\log_2{x} =\frac{4}{3}

    \log_2{x} = 1

    x=2
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  6. #6
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    3.

    3 \cos^2{A} - 2 \cos{A} - 1 = 0

    (3 \cos {A} + 1)(\cos{A} - 1)=0

    So either \cos{A} = -\frac{1}{3} or \cos{A}=1

    Solve each equation seperately
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  7. #7
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    thx alot...

    \log_2{x} + \log_8{x} =\frac{4}{3}

    \log_2{x} + \frac{\log_2{x}}{\log_2{8}} =\frac{4}{3}

    \log_2{x} + \frac{\log_2{x}}{3} =\frac{4}{3}

    \frac{4}{3}\log_2{x} =\frac{4}{3}---->how do you derive from this?care explain further?

    \log_2{x} = 1

    x=2

    3 \cos^2{A} - 2 \cos{A} - 1 = 0

    (3 \cos {A} + 1)(\cos{A} - 1)=0

    So either \cos{A} = -\frac{1}{3} or \cos{A}=1

    yeah i got 109.47deg,289.47deg,70.53 / 0deg,360
    am i right ? all anges???
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  8. #8
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    Hello, watsonmath!

    Glaysher's solution to #4 is absolutely correct.

    I was doing similar problems before learning the Base-change Formula.
    Because of the 'related' bases (2 and 8), we were taught to manipulate the logs.
    So here's my baby-talk version.


    4) Solve for x:\;\;\log_2x + \log_8x \:=\:\frac{4}{3}

    Let \log_8x = a. .Then: x = 8^a

    Take logs (base 2):
    . . \log_2x \:= \:\log_2(8^a) \:= \:\log_2(2^3)^a \:=\:\log_2(2)^{3a} \:= \:3a\log_22 \:=\:3a\cdot1\:=\:3a

    Hence, we have: . \log_2x\:=\:3a\:=\:3\log_8x


    The equation becomes: . 3\log_8x + \log_8x\:=\:\frac{4}{3}\quad\Rightarrow\quad4\log_  8x \,=  \frac{4}{3}\quad\Rightarrow\quad \log_8x \,= \,\frac{1}{3}<br />

    \text{Therefore: }\;x \,= \,8^{\frac{1}{3}}\quad\Rightarrow\quad \boxed{x \,= \,2}

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  9. #9
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    Quote Originally Posted by watsonmath

    \frac{4}{3}\log_2{x} =\frac{4}{3}---->how do you derive from this?care explain further?
    Adding fractions,
    \log_2 x+\frac{1}{3}\log_2 x
    \frac{3}{3}\log_2 x+\frac{1}{3}\log_2 x
    Thus,
    \frac{4}{3}\log_2 x
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  10. #10
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    thx... what about the trigo qns? any helper?
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  11. #11
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    Quote Originally Posted by watsonmath
    thx... what about the trigo qns? any helper?
    Which one, all were answered.
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  12. #12
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    Quote Originally Posted by ThePerfectHacker
    Which one, all were answered.
    here
    Find all angles between 0deg and 360deg inclusive which satisfy the equation : 3cossquartA - 2cosA - 1 = 0

    i got 109.47deg,289.47deg,70.53 / 0deg,360
    am i right ? all anges???
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  13. #13
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    Quote Originally Posted by watsonmath
    here
    Find all angles between 0deg and 360deg inclusive which satisfy the equation : 3cossquartA - 2cosA - 1 = 0

    i got 109.47deg,289.47deg,70.53 / 0deg,360
    am i right ? all anges???
    109 is wrong. Besides for that it is good.
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  14. #14
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    Quote Originally Posted by ThePerfectHacker
    109 is wrong. Besides for that it is good.
    ic....but
    A = cos-1 -1/3.... issnt that the ans for that??? which Glaysher asked me to solve???
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  15. #15
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    Hello, watsonmath!



    Find all angles [0^o,\,360^o]: . 3\cos^2A - 2\cos A - 1 \:= \:0

    i got 109.47, 289.47, 70.53, 0, 360. .Am i right?

    Your second angle is wrong. .(You don't add 180)
    . . And your third angle is also wrong.
    If \cos Ais negative, then Amust be in Quadrant 2 or Quadrant 3.


    The expression factors: . (3\cos A + 1)(\cos A - 1) \:=\:0

    And we have two equations to solve:

    . . 3\cos A + 1 \:= \:0\quad\Rightarrow\quad \cos A \:= \:-\frac{1}{3}\quad\Rightarrow\quad\boxed{A \:=\:109.46^o,\;250.53^o}

    . . \cos A - 1 \:=\:0\quad\Rightarrow\quad \cos A = 1\quad\Rightarrow\quad\boxed{A\:=\:0^o,\;360^o}

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