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Math Help - need some help with a take home sheet

  1. #1
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    need some help with a take home sheet

    1. use manual steps to solve Cos(x) Tan(x) + (3)^1/2 Cos(x) = 0


    2. use calculator methods: consider y=-3sin(x-3)+2 on .5radians < x < 5.5r

    find: max, min points, inflection, and give calculus reasons


    3. Antiderive:
    A. 3x/4+x^2 dx

    B. 3/(4-X^2)^1/2 dx

    C. 3/4+x^2 dx

    D. 3+2x/(1-x^2)^1/2


    4. if sin (x) = -3/7, x is in quad 3 and cos (y)= -2/5, y is in quad 2. calculate exact values for :
    a) sin (x-y) =
    b) tan (2x) =


    5. show steps to prove that:

    csc^2(x)tan(x)-cot(x) is equal to tan(x)csc^2(x)tan(x)-cot(x) = tan(x)



    thanks guys
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  2. #2
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    Re:

    RE:

    These are some good problems. You should show us what you have before we solve them...
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  3. #3
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Stuck686 View Post

    5. show steps to prove that:

    csc^2(x)tan(x)-cot(x) is equal to tan(x)csc^2(x)tan(x)-cot(x) = tan(x)
    is that exactly how the question was phrased?
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  4. #4
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    yes. and i have done nothing yet. i was planing on checking back here tomorrow night and seeing if the asnwers i had (will be oing tomorrow) are right
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  5. #5
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Stuck686 View Post
    1. use manual steps to solve Cos(x) Tan(x) + (3)^1/2 Cos(x) = 0
    First of all, the functions are cos(x) and tan(x), not Cos(x) and Tan(x). Yes, capital letters are important!

    You need to factor:
    cos(x)[tan(x) + sqrt{3}] = 0

    So either
    cos(x) = 0 ==> x = (pi)/2, 3(pi)/2 rad
    or
    tan(x) + sqrt{3} = 0

    tan(x) = -sqrt{3}

    x = 2(pi)/3, 5(pi)/3 rad

    So
    x = (pi)/2, 2(pi)/3, 5(pi)/3, 3(pi)/2 rad

    -Dan
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  6. #6
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Stuck686 View Post
    3. Antiderive:
    A. 3x/4+x^2 dx

    B. 3/(4-X^2)^1/2 dx

    C. 3/4+x^2 dx

    D. 3+2x/(1-x^2)^1/2
    A. Int[3x/4+x^2 dx] = (3/4)*Int[x dx] + Int[x^2 dx]

    = (3/4)*(1/2)x^2 + (1/3)x^3 + C


    B. Int[3/(4-X^2)^1/2 dx] = 3*Int[1/sqrt{4 - x^2} dx]

    Let y = 2x ==> dy = 2 dx

    Int[3/(4-X^2)^1/2 dx] = 3*Int[1/sqrt{4 - x^2} dx] = 3*Int[1/sqrt{4 - 4y^2} (dy/2)]

    = (3/2)*Int[(1/2)*1/sqrt{1 - y^2} dy]

    = (3/4)*Int[1/sqrt{1 - y^2} dy]

    Now let y = sin(t) ==> dy = cos(t) dt
    Int[3/(4-X^2)^1/2 dx] = (3/4)*Int[1/sqrt{1 - y^2} dy]

    = (3/4)*Int[1/sqrt{1 - sin^2(t)} * cos(t) dt]

    = (3/4)*Int[1/cos(t) * cos(t) dt]

    = (3/4)*Int[dt] = (3/4)*t + C = (3/4)*asn(y) + C

    = (3/4)*asn(2x) + C


    C. Int[3/4+x^2 dx] = 3/4*Int[dx] + Int[x^2 dx]

    = (3/4)*x + (1/2)x^3 + C


    D. Int[3+2x/(1-x^2)^1/2 dx] = 3*Int[dx] + Int[2x/sqrt{1 - x^2} dx]

    For the second integral let y = 1 - x^2 ==> dy = 2x dx
    Int[3+2x/(1-x^2)^1/2 dx] = 3*Int[dx] + Int[2x/sqrt{1 - x^2} dx]

    = 3*x + Int[1/sqrt{y} (-dy)] = 3x - (2)*sqrt{y} + C

    = 3x - 2*sqrt{1 - x^2} + C

    -Dan
    Last edited by topsquark; May 8th 2007 at 03:47 AM.
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  7. #7
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Stuck686 View Post
    4. if sin (x) = -3/7, x is in quad 3 and cos (y)= -2/5, y is in quad 2. calculate exact values for :
    a) sin (x-y) =
    b) tan (2x) =
    a) sin(x - y) = sin(x)cos(y) - sin(y)cos(x)

    Now, sin(y) = -sqrt{1 - cos^2(y)} <-- Negative since y is in QIII

    sin(y) = -sqrt{1 - (-2/5)^2} = -sqrt{1 - 4/25} = -sqrt{21/25}

    cos(x) = -sqrt{1 - sin^2(x)} <-- Negative since x is in QII

    cos(x) = -sqrt{1 - (-3/7)^2} = -sqrt{1 - 9/49} = -sqrt{40/49}

    So
    sin(x - y) = sin(x)cos(y) - sin(y)cos(x)

    = (-3/7)(-2/5) - (-sqrt{21/25})(-sqrt{40/49})

    = 21/35 - sqrt{(21*40)/(25*49)}

    = 21/35 - (2/35)*sqrt{210}

    = (21 - 2*sqrt{210})/35


    b. tan(2x) = sin(2x)/cos(2x) = 2*sin(x)*cos(x)/(2*cos^2(x) - 1)

    = 2*(-3/7)*(-sqrt{40/49})/(2*(-sqrt{40/49})^2 - 1)

    = (6*sqrt{40/49)/7)/(2*40/49 - 1)

    = (6*sqrt{40}/49)/(80/49 - 1)

    = (12*sqrt{10}/49)/(31/49) <-- Cancel the 49s

    = 12*sqrt{10}/31

    -Dan
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  8. #8
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Stuck686 View Post
    5. show steps to prove that:

    csc^2(x)tan(x)-cot(x) is equal to tan(x)csc^2(x)tan(x)-cot(x) = tan(x)
    This is not possible, which is why Jhevon was asking about it. You have multiplied the first term by a tan(x) and said it is somehow equal to the original. A simpler version of what you are saying:
    Show that x + 2 = 2x + 2 for all x.

    This simply isn't true. (For x = 0, yes, but not for all x.)

    -Dan
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  9. #9
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    oh i read this POORLY written worksheet wrong.
    the problem is:

    SHOW THAT
    csc^2(x)tan(x)-cot(x)= tan(x)
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  10. #10
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Stuck686 View Post
    oh i read this POORLY written worksheet wrong.
    the problem is:

    SHOW THAT
    csc^2(x)tan(x)-cot(x)= tan(x)
    Convert everything into sines and cosines:
    csc(x) = 1/sin(x)
    tan(x) = sin(x)/cos(x)
    cot(x) = cos(x)/sin(x)

    So we have to show:
    1/[sin(x)cos(x)] - cos(x)/sin(x) = sin(x)/cos(x)

    Add the fractions on the LHS:
    1/[sin(x)cos(x)] - cos^2(x)/[sin(x)cos(x)]

    = [1 - cos^2(x)]/[sin(x)cos(x)]

    = sin^2(x)/[sin(x)cos(x)]

    So as long as sin(x) is not equal to 0 (which we require anyway for csc(x) and cot(x) to exist)

    = sin(x)/cos(x)

    which is the RHS.

    -Dan
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  11. #11
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Stuck686 View Post
    oh i read this POORLY written worksheet wrong.
    the problem is:

    SHOW THAT
    csc^2(x)tan(x)-cot(x)= tan(x)
    Here:

    Tips:
    1) start on the most complicated side, it gives you more options to change things
    2) it is USUALLY better to change everything to sines and cosines. these are the ones students are more familiar with, so it's easier to see connections
    3) always consider working on both sides one at a time to bring them to the same thing. sometimes working on one side only to get the answer is too hard or even impossible in some sense

    EDIT: You're too quick for me Dan!...can i call you Dan?
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