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Math Help - Exact value help

  1. #1
    Newbie Neko Sama's Avatar
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    Unhappy Exact value help

    I don't understand how to do it well ><


    sin340 degrees?


    it says to use sum or difference of two angles formulas



    anyone help me and show me how u solve it??
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    Quote Originally Posted by Neko Sama View Post
    I don't understand how to do it well ><


    sin340 degrees?


    it says to use sum or difference of two angles formulas



    anyone help me and show me how u solve it??
    You can simplify as,
    \sin (360^o -20^o)=-\sin 20^o
    But I do not think you can find this throw half angle formula.
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  3. #3
    Newbie Neko Sama's Avatar
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    ????

    Quote Originally Posted by ThePerfectHacker View Post
    You can simplify as,
    \sin (360^o -20^o)=-\sin 20^o
    But I do not think you can find this throw half angle formula.
    hmmmm but the answer doesn't look like the way they did example problems

    ex) sin 150 degrees=sin(45^o +60^o) = sin 45^o cos 60^o + cos 45^o sin 60^o = the square root of 2/2 * 1/2+ square root of 2/2 * the square root of 3/2 = (the square root of 2 + the square root of 6)/ 4



    hmmm did me word my question wrong?
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    Quote Originally Posted by Neko Sama View Post
    hmmm did me word my question wrong?
    I understand what you are asking.
    The angles that you can find are (In first Quadrant),
    30,45,60
    Using half-angles on 30,60 we can also find,
    15,30
    Using Sum on 15 and 60 we can also find,
    75
    Thus,
    15,30,45,60,75
    Now can we use a linear combination of these to express 20?
    No, because these a multiples of 15 and 20 is not.
    So, you need a different approach.
    One way is to use the 1/3 angle formula.
    But I do not think the problem asks that.
    I think you got a wrong problem.
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  5. #5
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    I understand what you are asking.
    The angles that you can find are (In first Quadrant),
    30,45,60
    Using half-angles on 30,60 we can also find,
    15,30
    Using Sum on 15 and 60 we can also find,
    75
    Thus,
    15,30,45,60,75
    Now can we use a linear combination of these to express 20?
    No, because these a multiples of 15 and 20 is not.
    So, you need a different approach.
    One way is to use the 1/3 angle formula.
    But I do not think the problem asks that.
    I think you got a wrong problem.
    "1/3 angle formula?" Never heard of it. Would you mind sharing this particular secret of the Universe, O wise one?

    -Dan
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    Quote Originally Posted by topsquark View Post
    "1/3 angle formula?" Never heard of it. Would you mind sharing this particular secret of the Universe, O wise one?
    There is nothing amazing about that, I am sure you can derive it yourself.

    \cos 3x=\cos 2x\cos x-\sin 2x\sin x
    \cos 3x=(2\cos^2 x-1)\cos x-2\sin^2 x\cos x
    \cos 3x=2\cos^3 x-\cos x-2(1-\cos^2 x)\cos x
    \cos 3x=2\cos ^3x-\cos x-2\cos x-2\cos^3x
    \cos 3x=4\cos^3 x-3\cos x
    ---
    So let,
    x=\cos 15^o
    Thus,
    \frac{\sqrt{2}}{2}=4x^3-3x
    We need to find all the values of x that satisfy this equation.
    Thus,
    4x^3-3x=\frac{\sqrt{2}}{2}
    Divide through by 4,
    x^3-\frac{3}{4}x=\frac{\sqrt{2}}{8}
    ---

    Similarly, we can do the 1/5 angle formula.
    But that leads to a quintic which is not solvable by the premutation group of radicals. Which answers the question to why there is no \sin 1^o through radicals in this thread.

    Basically, not being able to express the unit angle through radicals is the same problem as not being able to construct some things through "Euclidean toys" (compass and straightedge).
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    There is nothing amazing about that, I am sure you can derive it yourself.

    \cos 3x=\cos 2x\cos x-\sin 2x\sin x
    \cos 3x=(2\cos^2 x-1)\cos x-2\sin^2 x\cos x
    \cos 3x=2\cos^3 x-\cos x-2(1-\cos^2 x)\cos x
    \cos 3x=2\cos ^3x-\cos x-2\cos x-2\cos^3x
    \cos 3x=4\cos^3 x-3\cos x
    ---
    So let,
    x=\cos 15^o
    Thus,
    \frac{\sqrt{2}}{2}=4x^3-3x
    We need to find all the values of x that satisfy this equation.
    Thus,
    4x^3-3x=\frac{\sqrt{2}}{2}
    Divide through by 4,
    x^3-\frac{3}{4}x=\frac{\sqrt{2}}{8}
    ---

    Similarly, we can do the 1/5 angle formula.
    But that leads to a quintic which is not solvable by the premutation group of radicals. Which answers the question to why there is no \sin 1^o through radicals in this thread.

    Basically, not being able to express the unit angle through radicals is the same problem as not being able to construct some things through "Euclidean toys" (compass and straightedge).
    I see. Yes, I have even used it, but I've never thought about it in terms of a formula (much less with a name.) Thanks.

    -Dan
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