Q12Mathematics I

In calculus, Taylor's theorem allows any infinitely differentiable function to be represented as an infinite sum of polynomial terms calculated from the values of its mathematical derivatives at a single specific point. The general formula for the Taylor series expansion of a function $f(x)$ centered around the point $x=a$ is rigorously defined as:

Step 1: Establishing the Function and Point

We are explicitly tasked with expanding the trigonometric function $f(x)=\sin x$ about the specific evaluation point $a=\pi /2$.

Step 2: Systematic Calculation of Derivatives

To construct the series, we must systematically calculate the continuous derivatives of $f(x)$ and then evaluate each of them exactly at $x=\pi /2$:

• Function (0th derivative): $f(x)=\sin x⟹f(\pi /2)=\sin (\pi /2)=1$
• 1st derivative: $f^{\prime }(x)=\cos x⟹f^{\prime }(\pi /2)=\cos (\pi /2)=0$
• 2nd derivative: $f^{\prime \prime }(x)=-\sin x⟹f^{\prime \prime }(\pi /2)=-\sin (\pi /2)=-1$
• 3rd derivative: $f^{\prime \prime \prime }(x)=-\cos x⟹f^{\prime \prime \prime }(\pi /2)=-\cos (\pi /2)=0$
• 4th derivative: $f^{\prime \prime \prime \prime }(x)=\sin x⟹f^{\prime \prime \prime \prime }(\pi /2)=\sin (\pi /2)=1$

This mathematical pattern $(1,0,-1,0)$ repeats infinitely for all subsequent higher-order derivatives.

Step 3: Constructing the Infinite Series

We meticulously substitute these evaluated derivative values back into the fundamental Taylor series formula:

Eliminating all the zero terms (which entirely removes all odd-powered terms) yields the final, perfectly elegant alternating series:

This derived expansion is actually mathematically identical to the Maclaurin series expansion for the cosine function, which completely makes sense analytically because of the fundamental trigonometric identity $\sin (x)=\cos (x-\pi /2)$.