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Math Help - Trigonometric Functions.. again

  1. #1
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    Post Trigonometric Functions.. again

    Hi...again.

    I know I'm a bother but can you guys help me again?

    I'm having trouble with these ones. Thanks!

    1. find the exact solutions algebraically in the interval [0, 2pi)
    sin 2x sin x = cos x. (2 is not an exponent on this one)

    2. verify identity algebraically
    cos^4 x - sin^4 x = cos 2x.

    3. find all solutions of the equation in the interval [0, 2pi)
    sin(x + pi/6) - sin(x - pi/6) = 1/2
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    Quote Originally Posted by cuteisa89 View Post

    2. verify identity algebraically
    cos^4 x - sin^4 x = cos 2x.
    Difference of two squares,

    (cos^2 x + sin^2 x)(cos^2 x - sin^2 x)

    The first factor is a Pythagorean Identity,
    Thus,

    cos^2 x - sin^2 x = cos 2x

    By double angle.
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  3. #3
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    Quote Originally Posted by cuteisa89 View Post
    3. find all solutions of the equation in the interval [0, 2pi)
    sin(x + pi/6) - sin(x - pi/6) = 1/2
    Add them (the sines) together to get,

    2*cos(2x)*sin(pi/3) = 1/2

    Thus,

    2*cos(2x)*(1/2) = (1/2)

    Thus,

    cos(2x) = 1/2
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  4. #4
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    Quote Originally Posted by cuteisa89 View Post
    1. find the exact solutions algebraically in the interval [0, 2pi)
    sin 2x sin x = cos x. (2 is not an exponent on this one)
    sin(2x)*sin(x) = cos(x)

    2*sin(x)*cos(x)*sin(x) = cos(x)

    2*sin^2(x)*cos(x) - cos(x) = 0

    cos(x)*[2*sin^2(x) - 1] = 0

    Thus
    cos(x) = 0 ==> x = (pi)/2, 3(pi)/2
    or
    2*sin^2(x) - 1 = 0

    2*sin^2(x) = 1

    sin^2(x) = 1/2

    sin(x) = (+/-)1/sqrt{2}

    x = (pi)/4, 3(pi)/4, 5(pi)/4, 7(pi)/4

    Thus
    x = (pi)/4, (pi)/2, 3(pi)/4, 5(pi)/4, 3(pi)/2, 7(pi)/4

    -Dan
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