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Math Help - need help - 3 various problems

  1. #1
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    need help - 3 various problems

    Hey guys, Im sorry to bothered you but Im kinda desperate and really need your help.

    Im preparing for math exam and Ive got a 3 various problem
    I know the right answers but Im not able to find out the method. I would really appreciate your help.

    1. if cos x = -2/5 than cos 2x = ?? (i know it is -17/25)
    or sin x = 3/5 and cos2x= ?? (7/25) but I have no clue how to do it.

    2. the number of X (X is from (pi, 2pi) which is valid for sin^2x-sinx=0
    (it is 0)

    or X is from (0, 2pi) for sin^3x+sinx=0
    (it is 1)

    3. the last one
    the imaginary part of the number
    (-1 -i)^8 is (0)
    or (1+i)^8 is (16)
    (-2 -2i)^8 is (2^12)
    (1+i)^32 is (0)

    Much appreciate your help and time!
    Last edited by mika619; June 5th 2010 at 09:06 AM.
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  2. #2
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    Quote Originally Posted by mika619 View Post
    Hey guys, Im sorry to bothered you but Im kinda desperate and really need your help.

    Im preparing for math exam and Ive got a 3 various problem
    I know the right answers but Im not able to find out the method. I would really appreciate your help.

    1. if cos x = -2/5 than cos 2x = ?? (i know it is -17/25)
    or sin x = 3/5 and cos2x= ?? (7/25) but I have no clue how to do it.
    Use \cos(2x) = 2 \cos^2x-1 = 1-2 \sin^2x

    Quote Originally Posted by mika619 View Post
    2. the number of X (X is from (pi, 2pi) which is valid for sin^2x-sinx=0
    (it is 0)

    or X is from (0, 2pi) for sin^3x+sinx=0
    (it is 1)
    Factor out \sin x

    Quote Originally Posted by mika619 View Post
    3. the last one
    the imaginary part of the number
    (-1 -i)^8 is (0)
    or (1+i)^8 is (16)
    (-2 -2i)^8 is (2^12)
    (1+i)^32 is (0)

    Much appreciate your help and time!
    -1-i = \sqrt{2} \: e^{i\frac{5\pi}{4}}
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  3. #3
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    ok, thanks for you help, anyway im dumb as hell so it wasn't actually really usefull for me...
    update: OK, it was helpful I got the first two... the 3th one is pure mess.

    I just found another I guess it is pretty simple but I can't solve it...

    <br />
 \frac{3 \sqrt{5} \ - 5 \sqrt{3}}{|\sqrt{3}-\sqrt{5}|} \
    thank you in advanced
    Last edited by mika619; June 5th 2010 at 11:48 AM.
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  4. #4
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    Quote Originally Posted by mika619 View Post
    ok, thanks for you help, anyway im dumb as hell so it wasn't actually really usefull for me...
    update: OK, it was helpful I got the first two... the 3th one is pure mess.
    -1-i = \sqrt{2} \: e^{i\frac{5\pi}{4}}

    (-1-i)^8 = \left(\sqrt{2} \: e^{i\frac{5\pi}{4}}\right)^8 = 16  \: e^{i\frac{8\times5\pi}{4}} = 16


    Quote Originally Posted by mika619 View Post
    I just found another I guess it is pretty simple but I can't solve it...

    <br />
 \frac{3 \sqrt{5} \ - 5 \sqrt{3}}{|\sqrt{3}-\sqrt{5}|} \
    thank you in advanced
    Since \sqrt{3} \leq \sqrt{5}

    \frac{3 \sqrt{5} \ - 5 \sqrt{3}}{|\sqrt{3}-\sqrt{5}|} =  \frac{3 \sqrt{5} \ - 5 \sqrt{3}}{\sqrt{5}-\sqrt{3}}

    Multiply numerator and denominator by \sqrt{5}+\sqrt{3}

    Expand and simplify
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  5. #5
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    I had the test today... I got 85% Anyway could you solve this for me
    (1+i)^33 I still don't have a clue. the options were I guess
    a)2^16
    b)2^7
    c)-2^16
    d)-2^17
    e)none of them
    thx

    nvm. google solve it
    (1 + i)^33 = 65 536 + 65 536 i
    Last edited by mika619; June 8th 2010 at 01:32 PM.
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  6. #6
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    Hello, mika619!

    Recall that: . i^4 \:=\:+1


    ( (1+i)^{33} \:=

    . . (a)\;2^{16} \qquad (b)\;2^7 \qquad(c)\;2^{16} \qquad (d)\;-2^{17 } \qquad (e)\;\text{none of these}

    Note that: . (1 + i)^2 \;=\;1 + 2i + i^2 \;=\;1 + 2i - 1

    . . So we have: . (1+i)^2 \;=\;2i



    Raise both sides to the 16th power:

    . . \bigg[(1+i)^2\bigg]^{16} \;=\;(2i)^{16}

    . . (1+i)^{32} \;=\;2^{16}\cdot i^{16} \;=\;65,536\,(i^4)^4 \;=\;65,536\,(+1)^4

    And we have: . (1+i)^{32} \;=\;65,\!536



    Multiply both side by (1+i):

    . . (1+i)^{32}\cdot(1+i) \;=\;65,\!536\,(1+i)



    Therefore: . (1+i)^{33} \;=\;65,\!536 + 65,\!536i

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  7. #7
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    Hello, mika619!

    Does #3 go like this?


    3. Find the imaginary part of these numbers:

    . . (a)\;(-1 -i)^8 \qquad (b)\;(1+i)^8 \qquad (c)\;<br />
(-2 -2i)^8 \qquad (d)\;(1+i)^{32}

    All four of these turn out to be real numbers,
    . . so their imaginary compoenents are all zero.


    (a)\;(-1-i)^8

    We find that: . (-1-i)^2 \:=\:1 + 2i + i^2 \:=\:1 + 2i - 1

    . . So we have: . (-1 - i)^2 \:=\:2i


    Raise both sides to the 4th power:

    . . \bigg[(1+i)^2\bigg]^4 \;=\;\bigg[2i\bigg]^4

    . . (-1-i)^8 \;=\;2^4\cdot i^4 \;=\;16\cdot1

    Therefore: . (-1-i)^8 \;=\;16



    (c)\;(-2-2i)^8

    We find that: . (-2-2i)^2 \;=\;4 + 8i + 4i^2 \;=\;4 + 8i - 4

    . . So we have: . (-2-2i)^2 \;=\;8i


    Raise both sides to the 4th power:

    . . \bigg[(-2-2i)^2\bigg]^4 \;=\;\bigg[8i\bigg]^4


    . . (-2-2i)^7 \;=\;8^4\cdot i^4 \;=\;4096\cdot1


    Therefore: . (-2-2i)^8 \;=\;4096


    Get the idea?

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