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    if anyone outthere knows the the answers to the following please post in ur replys......


    a) cos(@-10*) = cos 15*
    b) tan [email protected]=.4
    c) 2sin @tan=3

    @=theta *=degrees



    (btw this was the last question on the january 2004 paper so some of u would know the answers to this)
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    (Original post by gemstone)
    if anyone outthere knows the the answers to the following please post in ur replys......


    a) cos(@-10*) = cos 15*
    b) tan [email protected]=.4
    c) 2sin @tan=3

    @=theta *=degrees
    a) If cos (@-10*) = cos 15*
    then @-10 = 15
    => @ = 15 + 10
    =25

    b) tan [email protected] = 0.4
    [email protected] = tan^-1 (0.4)
    @ = [tan^-1 (0.4)]/2

    c) Don't quite understand what you've ritten for c....doesn't make sense.

    G

    correction to c)



    c) [email protected] [email protected]=3

    (this is correct)
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    (Original post by gemstone)
    correction to c)



    c) [email protected] [email protected]=3

    (this is correct)
    Fraid i can't figure it out....sorry..anyone else help out???

    G
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    (Original post by gzftan)
    Fraid i can't figure it out....sorry..anyone else help out???

    G
    [email protected]@=3

    [email protected]([email protected]/[email protected])=3 therefore:

    Sin^[email protected]=3/[email protected]
    1-Cos^[email protected]=3/[email protected]
    Cos^[email protected]/[email protected]=0

    Then solve as normally.
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    2sin^[email protected] = [email protected]

    Edit - just realised you meant 3/2 of [email protected] rather than 3/2 [email protected] Oh well, at least this others won't have similar confusion now
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    (Original post by Zapsta)
    2sin^[email protected] = [email protected]

    Edit - just realised you meant 3/2 of [email protected] rather than 3/2 [email protected] Oh well, at least this others won't have similar confusion now
    Yes, you can remove the denominator by multiplying the whole equation by 2. I had thought that people here are far more capeable than that.
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    Bhaal, sin^2(x) = 1-cos^2(x) so we actually have

    2cos^[email protected][email protected] = 0

    Then you factorise this a a quadratic in [email protected]; I used to find it easier to let [email protected] and then the equation becomes

    2y^2 + 3y - 2 = 0

    Which then looks easier to solve. But that is just a method I use.
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    (Original post by mikesgt2)
    Bhaal, sin^2(x) = 1-cos^2(x) so we actually have

    2cos^[email protected][email protected] = 0

    Then you factorise this a a quadratic in [email protected]; I used to find it easier to let [email protected] and then the equation becomes

    2y^2 + 3y - 2 = 0

    Which then looks easier to solve. But that is just a method I use.
    In my working out, I missed that step and automatically put that in.

    [email protected]@=3
    [email protected]([email protected]/[email protected])=3 therefore:
    Sin^[email protected]=3/[email protected]
    1-Cos^[email protected]=3/[email protected] <--see
    Cos^[email protected]/[email protected]=0
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    (Original post by Bhaal85)
    In my working out, I missed that step and automatically put that in.

    [email protected]@=3
    [email protected]([email protected]/[email protected])=3 therefore:
    Sin^[email protected]=3/[email protected]
    1-Cos^[email protected]=3/[email protected] <--see
    Cos^[email protected]/[email protected]=0
    What do you mean? As far as I am aware the quadratic is

    cos^[email protected]+3/[email protected]=0
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    (Original post by mikesgt2)
    What do you mean? As far as I am aware the quadratic is

    cos^[email protected]+3/[email protected]=0
    Times by 2 to get:

    2cos^[email protected][email protected] = 0 which is what you got, then factorise.

    2y^2+3y-2 = 0

    (2y-1)(y+2)=0

    Therefore [email protected]=1/2 and [email protected]=-2. The only real answer being [email protected]=1/2, therefore @=Cos^1(1/2). For which you should get two answers.

    wow u ppl r amazing...seriously u r....i never expected to get 9 replies to my question i thought i might have to look somewhere else.....wow i seriously am grateful thanx sooooooooooooooooooooooo much.

    cheers

    new question that i require help in sorry for sounding dumb but i need to make sure these r correct coz i havent got a mark scheme.

    f(x)=x^2 - kx + 9, where k is a constant

    (a) find the set of values of k for which the equation f(x) = 0 has no real solutions.




    Given that k=4
    (b) express f(x) in the form (x-p)^2 + q, where p and q are constants to b found



    (c) write down the minimum value of f(x) and the value of x for which this occurs
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    (Original post by gemstone)
    new question that i require help in sorry for sounding dumb but i need to make sure these r correct coz i havent got a mark scheme.

    f(x)=x^2 - kx + 9, where k is a constant

    (a) find the set of values of k for which the equation f(x) = 0 has no real solutions.
    Given that k=4
    (b) express f(x) in the form (x-p)^2 + q, where p and q are constants to b found
    (c) write down the minimum value of f(x) and the value of x for which this occurs
    b^2-4ac < 0
    k^2 - 36 < 0
    (k+6)(k-6) < 0
    -6 < k < 6

    f(x) = x^2 -4x +9
    = (x-2)^2 +9 - 4
    = (x-2)^2 +5

    5, at x=2

    thanx for part a and b...but how on earth do u get part (c)....can u plz explain to me how u get the 5 and x=2
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    Ok, this is something you will need to learn. When you complete the square on a function of x (NOT when x=0, in this case you are completing the square to FIND x), you get (x+a)^2+b = f(x)

    This is what you must remember: In this form, you can find the minimum point of the curve. the minimum point is the number outside the brackets (b), so if b is -5, the min. point of the curve you can possibly get is at y = -5. the corresponding x value is found bu putting the bracket to zero:

    x+a = 0
    x = -a

    Follow this through and you'll get the same result. Hope this helps

    thanx a lot that really helped....this is the first time i understand the principles of completing the square coz b4 i didnt have a clue..

    by the way wat yr r u in?

    i asked this b4 but i got 50 replies with many different answers....i hope to get a single answer this time....thanx loads


    [email protected] [email protected]=3
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    (Original post by gemstone)
    i asked this b4 but i got 50 replies with many different answers....i hope to get a single answer this time....thanx loads


    [email protected] [email protected]=3
    [email protected]@/[email protected] = 3

    2sin^[email protected] = [email protected]

    2(1-cos^[email protected]) = [email protected]

    2-2cos^[email protected] = [email protected]
    2cos^[email protected] + [email protected] - 2 = 0

    ([email protected])([email protected]+2 )

    [email protected] = 1/2 or - 2

    @ = 60, 300 between 0 and 360

    thanx a lot that reallly sorted so many things out
 
 
 
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