Find the Taylor polynomial of orders $0,1,2$, and 3 generated by $f$ at a. \[ f(x)=\cos x, a=\frac{\pi}{4} \]

Step 1 ：Given the function \(f(x) = \cos(x)\) and \(a = \frac{\pi}{4}\), we are asked to find the Taylor polynomials of orders 0, 1, 2, and 3.

Step 2 ：The Taylor polynomial of order n for a function f at a point a is given by: \[P_n(x) = f(a) + f'(a)(x-a) + \frac{f''(a)(x-a)^2}{2!} + \frac{f'''(a)(x-a)^3}{3!} + ... + \frac{f^n(a)(x-a)^n}{n!}\]

Step 3 ：We first find the derivatives of \(f(x) = \cos(x)\) up to the third order: \[f'(x) = -\sin(x),\ f''(x) = -\cos(x),\ f'''(x) = \sin(x)\]

Step 4 ：We then evaluate these derivatives at \(a = \frac{\pi}{4}\): \[f(a) = \frac{\sqrt{2}}{2},\ f'(a) = -\frac{\sqrt{2}}{2},\ f''(a) = -\frac{\sqrt{2}}{2},\ f'''(a) = \frac{\sqrt{2}}{2}\]

Step 5 ：We substitute these values into the Taylor polynomial formula to get the Taylor polynomials of orders 0, 1, 2, and 3: \[P_0(x) = \frac{\sqrt{2}}{2},\ P_1(x) = -\frac{\sqrt{2}}{2}(x - \frac{\pi}{4}) + \frac{\sqrt{2}}{2},\ P_2(x) = -\frac{\sqrt{2}}{4}(x - \frac{\pi}{4})^2 - \frac{\sqrt{2}}{2}(x - \frac{\pi}{4}) + \frac{\sqrt{2}}{2},\ P_3(x) = \frac{\sqrt{2}}{12}(x - \frac{\pi}{4})^3 - \frac{\sqrt{2}}{4}(x - \frac{\pi}{4})^2 - \frac{\sqrt{2}}{2}(x - \frac{\pi}{4}) + \frac{\sqrt{2}}{2}\]

Step 6 ：Final Answer: The Taylor polynomials of orders 0, 1, 2, and 3 generated by \(f(x)=\cos x\) at \(a=\frac{\pi}{4}\) are: \[\boxed{P_0(x) = \frac{\sqrt{2}}{2}},\ \boxed{P_1(x) = -\frac{\sqrt{2}}{2}(x - \frac{\pi}{4}) + \frac{\sqrt{2}}{2}},\ \boxed{P_2(x) = -\frac{\sqrt{2}}{4}(x - \frac{\pi}{4})^2 - \frac{\sqrt{2}}{2}(x - \frac{\pi}{4}) + \frac{\sqrt{2}}{2}},\ \boxed{P_3(x) = \frac{\sqrt{2}}{12}(x - \frac{\pi}{4})^3 - \frac{\sqrt{2}}{4}(x - \frac{\pi}{4})^2 - \frac{\sqrt{2}}{2}(x - \frac{\pi}{4}) + \frac{\sqrt{2}}{2}}\]