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  • 5 years ago

Using the first three terms of the Maclaurin's series, what is the value of cos(π/2

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  1. anonymous
    • 5 years ago
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    Do you mean cos(x/2)? If you do, its easier to start by differentiating the number of levels you need, in your case here, 3. f = cos(x/2) This derivative requires the chain rule. Let u = (x/2), then the chain rule says that we will have dcos(u)/du * d(x/2)/dx = -sin(u) * 1/2 Substitute back in and you get f^1 = -1/2sin(x/2) Do the same thing for f^2 and f^3 and I would recommend f^4 also. Now the Maclaurin series is the Taylor series at x = 0, so some of these terms will go away. The first term of the Maclaurin series is f(0), which in your case is cos(0/2) = cos(0) = 1 The second term goes away as sin(0) = 0. If you have done the other two derivatives you will see that the next term is -1/4cos(0/2) = -1/4 * (x^2/2!) The third term goes away again so it might be good to get another derivative in here. The fourth term comes out to be 1/16cos(0/2) = 1/16 * (x^4/4!) The first three terms of cos(x/2) is: 1 - (1/4)*(x^2/2!) + (1/16)*(x^4/4!) This is an alternating series and as the limit of x->0 this will come out to be 2. Hope this helps a little.

  2. anonymous
    • 5 years ago
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    \[\cos x = 1 -(x ^{2}/2!)+(x ^{4}/4!)\] where x = \[\pi/2\]

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