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Math Help - Double-Angle Formulas

  1. #1
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    Double-Angle Formulas

    Find the exact values of sin 2u, cos 2u, and tan 2u using the double angle formulas.:
    cos u = -2/3, pi/2 < u < pi

    So I know that the formulas are:
    sin 2u = 2 sin u cos u
    cos 2u = cos^2 u - sin^2 u
    tan 2u = (2 tan u) / (1 - tan^2 u)

    Now I am having trouble figuring out what sin u is. I believe that it is
    sqrt(3), but I am not sure.
    If I am right, the first formula would be:
    sin 2u = (2) (sqrt3) (-2/3)
    but I'm not sure how to multiply a square root with a fraction.

    Any help is appreciated.
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    Quote Originally Posted by iluvmathbutitshard View Post
    Find the exact values of sin 2u, cos 2u, and tan 2u using the double angle formulas.:
    cos u = -2/3, pi/2 < u < pi

    So I know that the formulas are:
    sin 2u = 2 sin u cos u
    cos 2u = cos^2 u - sin^2 u
    tan 2u = (2 tan u) / (1 - tan^2 u)

    Now I am having trouble figuring out what sin u is. I believe that it is
    sqrt(3), but I am not sure.
    If I am right, the first formula would be:
    sin 2u = (2) (sqrt3) (-2/3)
    but I'm not sure how to multiply a square root with a fraction.

    Any help is appreciated.
    nope, that's not the right value for sin(u). there are several ways you can get to sin(u). Two conventional ways are:

    Use the fact that \sin^2 u + \cos^2 u = 1

    Solve for \sin u. Note that we are in the second quadrant so you will need to take the positive square root. As sine is positive in that quadrant.


    Another way: Draw a right-triangle with an acute angle u. The adjacent side to u should be labeled 2 while the hypotenuse should be labeled 3. Use Pythagoras' Theorem to find the opposite side and use sine = opposite/hypotenuse
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    So sin u would be sqrt5/3?

    If so, how would I multiply this by -2/3?

    (Thank you for your help)
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    Quote Originally Posted by iluvmathbutitshard View Post
    So sin u would be sqrt5/3?
    yes, \sin u = \frac {\sqrt 5}3

    If so, how would I multiply this by -2/3?

    (Thank you for your help)
    the same way you multiply any two fractions... multiply the numerators together to get the new numerator and the denominators together to get the new denominator. \frac {\sqrt 5}3 \cdot \frac {-2}3 = \frac {-2 \sqrt 5}9
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  5. #5
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    Ok, great

    SO then for cos 2u:
    cos^2 u - sin^2 u
    (4/9) - (5/9) = (-1/9)

    Is this correct?
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    Quote Originally Posted by iluvmathbutitshard View Post
    Ok, great
    is it really? What was your answer for sin(2u)?

    SO then for cos 2u:
    cos^2 u - sin^2 u
    (4/9) - (5/9) = (-1/9)

    Is this correct?
    yup
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  7. #7
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    My answer for sin u was
    (-4sqrt5) / (9)

    And one last thing, is my tan correct:
    (2 tan u) / (1 - tan^2 u)
    (2 (-sqrt5/2)) / (1 - (5/4))

    So this would simplify, I think, to
    [(-2 sqrt5) / (2)] / [-1/4]
    Is this correct?
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    Quote Originally Posted by iluvmathbutitshard View Post
    My answer for sin u was
    (-4sqrt5) / (9)
    ok, good.

    And one last thing, is my tan correct:
    (2 tan u) / (1 - tan^2 u)
    (2 (-sqrt5/2)) / (1 - (5/4))

    So this would simplify, I think, to
    [(-2 sqrt5) / (2)] / [-1/4]
    Is this correct?
    that's not simplified

    for this last question, I would use the fact that \tan 2 u = \frac {\sin 2u}{\cos 2u}

    just divide the first two answers you got...
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  9. #9
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    So the answer would be 4 sqrt5?
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    Quote Originally Posted by iluvmathbutitshard View Post
    So the answer would be 4 sqrt5?
    yup
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  11. #11
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    Thank you so much for your help.
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