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Thread: 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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    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.

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

    $\displaystyle \cos(7\pi/12)$
    Next convert to degree if that is easier for thee,
    $\displaystyle \cos(7\cdot 180^o/12)=\cos(7\cdot 15^o)=\cos(105^o)$
    Next you can express this angle as,
    $\displaystyle \cos(180^o-75^o)=-\cos 75^o=-\sin 15^o$
    Now since,
    $\displaystyle \cos 30^o=\frac{\sqrt{3}}{2}$
    That means,
    $\displaystyle \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:
    $\displaystyle -\frac{\sqrt{3}-1}{2} \cdot \frac{\sqrt{2}}{4} \approx -0.12941$

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

    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:

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

    what he did say was:

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

    and that:

    $\displaystyle
    \sin(15^{\circ})=\frac{\sqrt{2-\sqrt{3}}}{2}
    $

    so:

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

    RonL
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    Re:

    Ok thanks Captain Black
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