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dpflan

  • 3 years ago

What is the linear approximation for sin(x) at x=0? How can I use the answer the approximate the values of sin(.01), sin(.1) and sin(1)?

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  1. JoanLu
    • 3 years ago
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    y=x ; \[\sin(.01)\approx.01\] \[\sin(.1)\approx.1\]\[\sin(1)\approx1\]

  2. JoanLu
    • 3 years ago
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    y=x ; \[\sin(.01)\approx.01\] \[\sin(.1)\approx.1\]\[\sin(1)\approx1\]

  3. kh83
    • 3 years ago
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    Let's see sin(0.01) = 0.00999, relative error = 0.002%, very good sin(0.1) = 0.099, error 0.1% that's good sin(1) = 0.8415, error = 16 % - that's unacceptable but 1 radian is pretty big angle

  4. l
    • 3 years ago
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    It depends on what type of equation/formulation you are using. For example if y = 1/sin(x). There's a great difference between sin0, sin0.1 and sin 1. As a numerical analysis method you can look up Newton's Approximation Formula: http://en.wikipedia.org/wiki/Newton's_method *

  5. hburgiel
    • 3 years ago
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    The linear approximation for sin(x) at x=0 is just the equation of the tangent line to the graph of sin(x) at x=0. So \[\sin(x) \approx x.\]

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