sine approximation using a loop
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The link above should have my work on it could someone help explain what I am doing wrong.
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Walter Roberson
le 1 Oct 2020
Modifié(e) : Walter Roberson
le 1 Oct 2020
x = pi/2;
That is overriding the user input about the value to find the sine of.
function [s] = prob3_2(x,N)
You never use the user input N
n = 1;
you have a variable named n which is confusing considering the input named N
while error >= 1*(10^-3);
You are proceeding until your error is in a certain range, but the problem requirement says
% takes a scalar angle measure x (in radians) and estimates sin(x) using
% the first N terms of an alternating series
so you are not intended to go until error is a particular range: you are to go until you have used a particular number of terms.
error = abs((sin(x)-s)/sin(x))*100;
Is error a percentage, or is an absolute value?
What about the cases where sin(x) = 0 ? You would be dividing by 0, which is not going to give you a useful error measure.
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