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Math Help - Approximation of Sin Function

  1. #1
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    Approximation of Sin Function

    I cannot figure out how to work this problem:

    Use differentials (or equivalently, a linear approximation) to approximate as follows: Let and find the equation of the tangent line to at a "nice" point near . Then use this to approximate .

    Please help!
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Quiz0n View Post
    I cannot figure out how to work this problem:

    Use differentials (or equivalently, a linear approximation) to approximate as follows: Let and find the equation of the tangent line to at a "nice" point near . Then use this to approximate .

    Please help!
    recall by the linear approximation formula: f(x) \approx f(a) + f'(a)(x - a)

    where f(x) is the value you want to find and a is a value very close to x for which we do know what f(a) and f'(a) are.



    Here, it is almost obvious that a = 45^o, since you should know what \sin \left( 45^o \right) is. so just plug the numbers into the formula, and it should be pretty straight forward from there...of course, if it isn't i don't expect you to act as if it is, say something
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    I can get up to finding the equation of the tangent line, but I don't know where to go from there to find the approximation of sin(40). Am I supposed to just plug 40 in as "x" in the linearization?
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Quiz0n View Post
    I can get up to finding the equation of the tangent line, but I don't know where to go from there to find the approximation of sin(40). Am I supposed to just plug 40 in as "x" in the linearization?
    yes, as i said, x = 40 and a = 45. plug those into the formula
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  5. #5
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    I'm setting x=40 and a=45 into the equation:

    f(x) = sin(x) + cos(x) (x-a)

    Furthermore:

    f(40) = (2^(1/2) / 2) + (2^(1/2) / 2) (40 - 45)

    This equation gives a value of -2.82843, which is being denied by my assignment program, what am I doing wrong?
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Quiz0n View Post
    I'm setting x=40 and a=45 into the equation:

    f(x) = sin(x) + cos(x) (x-a)

    Furthermore:

    f(40) = (2^(1/2) / 2) + (2^(1/2) / 2) (40 - 45)

    This equation gives a value of -2.82843, which is being denied by my assignment program, what am I doing wrong?
    oh, that's because when using trig functions, this formula does not work well with degrees. try with x = \frac {2 \pi}9 and a = \frac {\pi}4 and you will get the right answer. with degrees, the difference is too great, we need a relatively small distance between the points (since we are using a straight line to approximate a curve), we obtain that by working in radians, the difference between the radian values is about 0.0278, which is not as big as 5. it is more ideal.

    (i hope you remember how to change degrees to radians)
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  7. #7
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    Quote Originally Posted by Jhevon View Post
    yes, as i said, x = 40 and a = 45. plug those into the formula
    Quote Originally Posted by Quiz0n View Post
    I'm setting x=40 and a=45 into the equation:

    f(x) = sin(x) + cos(x) (x-a)

    Furthermore:

    f(40) = (2^(1/2) / 2) + (2^(1/2) / 2) (40 - 45)

    This equation gives a value of -2.82843, which is being denied by my assignment program, what am I doing wrong?
    That's because when you take a derivative of a trig function it is assumed the argument is in radians. So convert your 40 and 45 degrees to radians and it will work.

    -Dan
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by topsquark View Post
    That's because when you take a derivative of a trig function it is assumed the argument is in radians. So convert your 40 and 45 degrees to radians and it will work.

    -Dan
    yes, that was my bad , i corrected myself though
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  9. #9
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    Thanks a ton guys, you're really a big help with this.
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