Next, evaluate the derivatives at the center x = f(x) = cos(x) = f(n/4) = 2 √2 f'(x)=-sin(x) = f'(n/4) = f(x) = cos(x) = f(n/4) = 22 22 √2 f(x)=sin(x) => ""(n/4) = 22 (4)(x) = cos(x) -> (4)(π/4) √2 2 Π 1 √2 1 1 √2 Step 3 Substitute the derivative values in the Taylor series centered at x= cos(x)= XT 2 πT and simplify. (Hint: Note that (-1)(n+1)/2 = 1, −1, −1, 1, 1, -1, -1, 1 ...) 4 7) - √2 (1x-(1/4)]²) + √2 (Ex - (n/4)]³) 3! [x-(n/4)]4 = -ΣΟ 7=0 Submit Skip (you cannot come back) Submit Answer DETAILS LARCALC12 9.10.009.

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