How to find an Approximate Polynomial using Perceptron

Question

I want to approximate cos(x) from 0 to pi/4 using a quadratic polynomial.

I believe I could train a perceptron using (say 1000) points in my range, with training inputs (x^2, x, 1) and with training outputs sigmoid(cos(x)). Then, the resulting weights of the neuron would be the coefficients of the polynomial (namely w1x^2 + w2x + w3)

Here is my attempt (but it is not converging)

import numpy
import matplotlib.pyplot as plt
import random
import math
#Generate random points in range
R = [random.uniform(0, math.pi/4) for i in range(1000)]

def sigmoid(x):
    return 1.0/(1.0 + numpy.exp(-x))

def sigmoid_prime(x):
    return sigmoid(x)*(1 - sigmoid(x))

#Generate training inputs
training_inputs = numpy.ones((len(R), 3))
for index, x in enumerate(R):
    training_inputs[index, 0] = x**2
    training_inputs[index, 1] = x

#Generate training outputs
training_outputs = numpy.array([sigmoid(math.cos(i)) for i in R]).reshape(len(R), 1)

#Arbitrary weights
weights = numpy.ones((3, 1))

#Neural Network
def train_nn(training_inputs, training_outputs, initial_weights, niter, errors_data):
    w = initial_weights
    for ii in range(niter):
        #forward propagation
        outputs = sigmoid(numpy.dot(training_inputs, w))
        #backward propagation
        errors = training_outputs - outputs
        deltaw = errors*sigmoid_prime(outputs)
        deltaw = numpy.dot(training_inputs.T, deltaw)
        w += deltaw
        #save errors
        errors_data[ii] = errors.reshape(len(R),)
    return outputs, w


NITER = 5000
errors = numpy.zeros((NITER, len(R)))
outputs, weights = train_nn(training_inputs, training_outputs, weights, NITER, errors)

#coefficients
print(weights)