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Math Help - Trig Problem

  1. #1
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    Trig Problem

    The problem is:

    cos(-7pi/12)





    I used the half angle formula to get: -(sqrt(3)-1)*sqrt(2)/(4)


    Somehow or another the computer system isn't taking my answer.

    Thanks for the help!
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  2. #2
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    Quote Originally Posted by qbkr21 View Post
    The problem is:

    cos(-7pi/12)





    I used the half angle formula to get: -(sqrt(3)-1)*sqrt(2)/(4)


    Somehow or another the computer system isn't taking my answer.

    Thanks for the help!
    First drop the negative, cosine is an even function.

    \cos(7\pi/12)
    Next convert to degree if that is easier for thee,
    \cos(7\cdot 180^o/12)=\cos(7\cdot 15^o)=\cos(105^o)
    Next you can express this angle as,
    \cos(180^o-75^o)=-\cos 75^o=-\sin 15^o
    Now since,
    \cos 30^o=\frac{\sqrt{3}}{2}
    That means,
    \sin 15^o=\sqrt{\frac{1-\frac{\sqrt{3}}{2}}{2}}=\frac{\sqrt{2-\sqrt{3}}}{2}
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    Re:

    This seems to be the same answer I got, and it still doesn't want to fit into the system...Is there another way you can express this?

    Thanks!
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  4. #4
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    Quote Originally Posted by qbkr21 View Post
    The problem is:

    cos(-7pi/12)





    I used the half angle formula to get: -(sqrt(3)-1)*sqrt(2)/(4)


    Somehow or another the computer system isn't taking my answer.

    Thanks for the help!
    Quote Originally Posted by ThePerfectHacker View Post
    First drop the negative, cosine is an even function.

    \cos(7\pi/12)
    Next convert to degree if that is easier for thee,
    \cos(7\cdot 180^o/12)=\cos(7\cdot 15^o)=\cos(105^o)
    Next you can express this angle as,
    \cos(180^o-75^o)=-\cos 75^o=-\sin 15^o
    Now since,
    \cos 30^o=\frac{\sqrt{3}}{2}
    That means,
    \sin 15^o=\sqrt{\frac{1-\frac{\sqrt{3}}{2}}{2}}=\frac{\sqrt{2-\sqrt{3}}}{2}
    Quote Originally Posted by qbkr21 View Post
    This seems to be the same answer I got, and it still doesn't want to fit into the system...Is there another way you can express this?

    Thanks!
    Unless you changed your original answer you had:
    -\frac{\sqrt{3}-1}{2} \cdot \frac{\sqrt{2}}{4} \approx -0.12941

    ThePerfectHacker had:
    \frac{\sqrt{2 - \sqrt{3}}}{2} \approx 0.258819

    So the answers are not the same.

    -Dan
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    Re:

    Here are the directions the computer system is still not taking it:

    Use a sum or difference formula or a half angle formula to determine the value of the trigonometric functions. Give exact answers. Do not use decimal numbers. The answer should be a fraction or an arithmetic expression. If the answer involves a square root it should be enter as sqrt; e.g. the square root of 2 should be written as sqrt(2);



    Thanks!
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    Quote Originally Posted by qbkr21 View Post
    Here are the directions the computer system is still not taking it:

    Use a sum or difference formula or a half angle formula to determine the value of the trigonometric functions. Give exact answers. Do not use decimal numbers. The answer should be a fraction or an arithmetic expression. If the answer involves a square root it should be enter as sqrt; e.g. the square root of 2 should be written as sqrt(2);



    Thanks!
    You should enter:
    sqrt(2 - sqrt(3))/2

    or

    (sqrt(2 - sqrt(3)))/2

    The first one should do the trick, the second is being extra careful.

    -Dan
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  7. #7
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    Quote Originally Posted by qbkr21 View Post
    Here are the directions the computer system is still not taking it:

    Use a sum or difference formula or a half angle formula to determine the value of the trigonometric functions. Give exact answers. Do not use decimal numbers. The answer should be a fraction or an arithmetic expression. If the answer involves a square root it should be enter as sqrt; e.g. the square root of 2 should be written as sqrt(2);



    Thanks!
    Read what ImPerfectHacker wrote carefully.

    He did not say that:

    \cos(105^{\circ})=\sin(15^{\circ})

    what he did say was:

    \cos(105^{\circ})=-\sin(15^{\circ})

    and that:

    <br />
\sin(15^{\circ})=\frac{\sqrt{2-\sqrt{3}}}{2}<br />

    so:

    \cos(105^{\circ})=-\frac{\sqrt{2-\sqrt{3}}}{2}

    RonL
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  8. #8
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    Re:

    Ok thanks Captain Black
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