Question

a graphing calculator is recommended.
find the taylor polynomial ( t _ { 3 } ( x ) ) for the function ( f ) centered at the number ( a ).
( f ( x ) = cos ( x ), quad a = \frac { pi } { 2 } )
( t _ { 3 } ( x ) = )
graph ( f ) and ( t _ { 3 } ) on the same screen.

resources
awesome job!

Explanation:

Step1: Recall the Taylor polynomial formula

The Taylor polynomial \(T_{n}(x)\) of degree \(n\) for a function \(f(x)\) centered at \(a\) is given by \(T_{n}(x)=\sum_{k = 0}^{n}\frac{f^{(k)}(a)}{k!}(x - a)^{k}\), where \(f^{(k)}(x)\) is the \(k\) - th derivative of \(f(x)\).
For \(n = 3\), \(T_{3}(x)=f(a)+f^{\prime}(a)(x - a)+\frac{f^{\prime\prime}(a)}{2!}(x - a)^{2}+\frac{f^{(3)}(a)}{3!}(x - a)^{3}\)

Step2: Find the derivatives of \(f(x)=\cos(x)\)

• \(f(x)=\cos(x)\), so \(f^{\prime}(x)=-\sin(x)\), \(f^{\prime\prime}(x)=-\cos(x)\), \(f^{(3)}(x)=\sin(x)\)
• Given \(a=\frac{\pi}{2}\)
• \(f(\frac{\pi}{2})=\cos(\frac{\pi}{2}) = 0\)
• \(f^{\prime}(\frac{\pi}{2})=-\sin(\frac{\pi}{2})=- 1\)
• \(f^{\prime\prime}(\frac{\pi}{2})=-\cos(\frac{\pi}{2}) = 0\)
• \(f^{(3)}(\frac{\pi}{2})=\sin(\frac{\pi}{2})=1\)

Step3: Substitute into the Taylor polynomial formula

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Answer:

\(T_{3}(x)=-(x-\frac{\pi}{2})+\frac{1}{6}(x - \frac{\pi}{2})^{3}\)