Question

What is the approximation for the value of cos(1) obtained by using the fourth-degree Taylor polynomial for cos x about x = 0 ? 1 A 1 + 1 64 B 1 + 1 384 с. 1 4 + 1o 1 1 1 D 1 + 36 4

Answer 1

`\cos(1)\approx0.54167`

Explanation:

`f(x)=f(a)+f'(a)(x-a)+(f''(a)(x-a)^2)/(2!)+(f''(a)(x-a)^3)/(3!)+...+(f^n(a)(x-a)^n)/(n!)`

`f(0)=\cos(0)=1\\f'(0)=-\sin(0)=0\\f''(0)=-\cos(0)=-1\\f'''(0)=\sin(0)=0\\f^4(0)=\cos(0)=1`

`f(x)=f(0)+f'(0)(x-0)+(f''(0)(x-0)^2)/(2!)+(f''(0)(x-a)^3)/(3!)+(f^4(0)(x-0)^4)/(4!)\\\\f(x)=1-(x^2)/(2)+(x^4)/(24)\\\\\cos(1)\approx1-(1^2)/(2)+(1^4)/(24)=0.54167`