Q.Expand $\\sin x$ in ascending powers of $x-\\dfrac{\\pi}{4}$ upto three non-zero terms.

Taylor's Series. If $f(x)$ is infinitely differentiable at $x=a$, then $f$ can be expanded, in an interval around $a$, as a power series in $(x-a)$:

Why the coefficients are $\dfrac{f^{(n)}(a)}{n!}$. Writing $f(x)=A_0+\sum_{n\ge1}A_n(x-a)^n$ and substituting $x=a$ gives $A_0=f(a)$. Differentiating once and substituting $x=a$ gives $A_1=f'(a)$; differentiating again gives $A_2=\dfrac{f''(a)}{2!}$; and in general, differentiating $k$ times and substituting $x=a$ isolates $A_k=\dfrac{f^{(k)}(a)}{k!}$ — every other term in the sum still carries a positive power of $(x-a)$ and vanishes at $x=a$.

Maclaurin's Series is simply the special case $a=0$:

Standard Maclaurin expansions worth knowing by heart (each derived by repeatedly differentiating and evaluating at $0$):

Working procedure for a given function.

1. Compute $f,f',f'',f''',\ldots$ (as many derivatives as the number of required terms) and evaluate each at the expansion point ($x=a$ for Taylor, $x=0$ for Maclaurin) — a small table (as in Table 7.2–7.4 of the text) keeps this organised.
2. Substitute each value into $\dfrac{f^{(n)}(a)}{n!}(x-a)^n$ and sum.
3. When the request is "about $x=a$" for $a\ne0$, always expand in powers of $(x-a)$, never in powers of $x$ directly. …