Find the Taylor series of f(x) = sin(x) at a = Pi/3.
To find the Taylor series of \(f(x) = \sin(x)\) at \(a = \frac{\pi}{3}\), we can use the Taylor series expansion formula. The general formula for the Taylor series expansion of a function \(f(x)\) centered at \(a\) is:
\[f(x) = f(a) + f'(a)(x - a) + \frac{f''(a)}{2!}(x - a)^2 + \frac{f'''(a)}{3!}(x - a)^3 + \ldots\]
In this case, since \(f(x) = \sin(x)\), let's find the derivatives of \(\sin(x)\) up to the desired order:
\[f(x) = \sin(x)\]
\[f'(x) = \cos(x)\]
\[f''(x) = -\sin(x)\]
\[f'''(x) = -\cos(x)\]
Next, let's substitute the values of \(f(a)\), \(f'(a)\), \(f''(a)\), \(f'''(a)\), and \(a\) into the Taylor series expansion formula:
\[f(\frac{\pi}{3}) = \sin(\frac{\pi}{3}) = \sqrt{3}/2\]
\[f'(\frac{\pi}{3}) = \cos(\frac{\pi}{3}) = 1/2\]
\[f''(\frac{\pi}{3}) = -\sin(\frac{\pi}{3}) = -\sqrt{3}/2\]
\[f'''(\frac{\pi}{3}) = -\cos(\frac{\pi}{3}) = -1/2\]
Now we can substitute these values into the Taylor series expansion formula:
\[\sin(x) = \frac{\sqrt{3}}{2} + \frac{1}{2}(x - \frac{\pi}{3}) - \frac{\sqrt{3}}{2}\frac{(x - \frac{\pi}{3})^2}{2!} - \frac{1}{2}\frac{(x - \frac{\pi}{3})^3}{3!} + \ldots\]
Simplifying further is possible, but presenting the infinite series can be concise enough for most purposes. That's the Taylor series expansion of \(f(x) = \sin(x)\) at \(a = \frac{\pi}{3}\).
sin(x-π/3) = sin(π/3) + 1/1! cos(π/3)(x-π/3) - 1/2! sin(π/3)(x-π/3)^2 ...
= √3/2 + 1/2 (x-π/3) - √3/4 (x-π/3)^2 ...