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Math Help - Simplification

  1. #1
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    Simplification

    I have been trying to figure out if this problem could be simplified more.
    sqrt(cos^2(t) cos(2 sin(t))-sin^2(t))

    Thanks
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by alexgo View Post
    I have been trying to figure out if this problem could be simplified more.
    sqrt(cos^2(t) cos(2 sin(t))-sin^2(t))

    Thanks
    as in \displaystyle \sqrt{\cos^2 t \cdot \cos (2 \sin t) - \sin^2 t} ? Is this right?
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    Quote Originally Posted by Jhevon View Post
    as in \displaystyle \sqrt{\cos^2 t \cdot \cos (2 \sin t) - \sin^2 t} ? Is this right?
    yup
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by alexgo View Post
    yup
    well, not really. i don't see any simplification that would be worth the effort.

    why do you want to do this? what was the original problem?
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    Quote Originally Posted by Jhevon View Post
    well, not really. i don't see any simplification that would be worth the effort.

    why do you want to do this? what was the original problem?
    Original Problem is actually a Calc 3 problem. I was looking for Tangent vector of r(t)=sin(sint) i + cos(sint) j + Cost k
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  6. #6
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by alexgo View Post
    Original Problem is actually a Calc 3 problem. I was looking for Tangent vector of r(t)=sin(sint) i + cos(sint) j + Cost k
    oh! you mean the unit tangent vector? i suppose you were trying to simplify |r'(t)|

    in that case, you're way off. your |r'(t)| is incorrect. for the correct one, it is 1.

    Hint: you made a sign error. there is a minus (at least one) that should be a plus in what you did.
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    Quote Originally Posted by Jhevon View Post
    oh! you mean the unit tangent vector? i suppose you were trying to simplify |r'(t)|

    in that case, you're way off. your |r'(t)| is incorrect. for the correct one, it is 1.

    Hint: you made a sign error. there is a minus that should be a plus in what you did.
    AHHH, so when I square I take the negative with it, right?
    for r'(t)=cost \cdot cos(sint) i - cost \cdot sin(sint) j- sint k
    for |r'(t)|=\sqrt{(cost \cdot cos(sint))^2 +(-cost \cdot sin(sint))^2 +(-sint)^2}
    |r'(t)|=\sqrt{cos^2t \cdot (cos^2(sint) + sin^2(sint)) +sin^2t}
    |r'(t)|=\sqrt{cos^2t \cdot (1) +sin^2t}
    |r'(t)|=1
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  8. #8
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by alexgo View Post
    AHHH, so when I square I take the negative with it, right?
    for r'(t)=cost \cdot cos(sint) i - cost \cdot sin(sint) j- sint k
    for |r'(t)|=\sqrt{(cost \cdot cos(sint))^2 +(-cost \cdot sin(sint))^2 +(-sint)^2}
    |r'(t)|=\sqrt{cos^2t \cdot (cos^2(sint) + sin^2(sint)) +sin^2t}
    |r'(t)|=\sqrt{cos^2t \cdot (1) +sin^2t}
    |r'(t)|=1
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