Question

Using the values, cos(1.10000) = 0.453598 and cos(1.30000) = 0.267502


Using the values, cos(1.10000) = 0.453598 and cos(1.30000) = 0.267502, find an approximate value of cos(1.25000) using Lagrange Interpolation.


 The recipe for Lagrange Interpolation to fit an nth order polynomial to n+1 data points is:


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Using the values, cos(1.10000) = 0.453598 and cos(1.30000) = 0.267502
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