Answer
a. $a=1, b=0, c=-1/2$ and $g(x)=1-x^2/2$
b. $b=cos(a), c=sin(a)$ and $g(x)=cos(a)\ sin(x)+sin(a)\ cos(x)$
c. In part (a) $f'''(0)=g'''(0)$, but $g^{(4)}(x)\ne f^{(4)}(0)$
In part (b) $f'''(0)=g'''(0)$, and $g^{(4)}(x)= f^{(4)}(0)$
Work Step by Step
a. Step 1. Given the functions $f(x)=cos(x)$ and $g(x)=a+bx+cx^2$, we have $f'(x)=-sin(x), f''(x)=-cos(x)$ and $g'(x)=b+2cx, g''(x)=2c$.
Step 2. Using the first condition $f(0)=g(0)$, we have $a=1$
Step 3. Using the second condition $f'(0)=g'(0)$, we have $b=0$.
Step 4. Using the third condition $f''(0)=g''(0)$, we have $2c=-1$. or $c=-1/2$.
Step 5. We conclude that $a=1, b=0, c=-1/2$ and $g(x)=1-x^2/2$
b. Step 1. Given the functions $f(x)=sin(x+a)$ and $g(x)=b\ sin(x)+c\ cos(x)$, we have $f'(x)=cos(x+a), f''(x)=-sin(x+a)$ and $g'(x)=b\ cos(x)-c\ sin(x), g''(x)=-b\ sin(x)-c\ cos(x)$.
Step 2. Using the first condition $f(0)=g(0)$, we have $sin(a)=c$
Step 3. Using the second condition $f'(0)=g'(0)$, we have $cos(a)=b$.
Step 4. We conclude that $b=cos(a), c=sin(a)$ and $g(x)=cos(a)\ sin(x)+sin(a)\ cos(x)=sin(x+a)=f(x)$
c. Step 1. In part (a), $f'''(x)=-sin(x)=f'(x), f^{(4)}(x)=-cos(x)=f''(x)$ which indicates repetitions.
Step 2. In part (a) $g'''(x)=g^{(4)}(x)=0$, thus $f'''(0)=g'''(0)$, but $g^{(4)}(x)\ne f^{(4)}(0)$
Step 3. In part (b) $f'''(x)=-cos(x+a)=-f'(x), f^{(4)}(x)=sin(x+a)=-f''(x)$
Step 4. In part (b) $g'''(x)=-b\ cos(x)+c\ sin(x)=-g'(x), g^{(4)}(x)=b\ sin(x)+c\ cos(x)=g(x)$ (back to the start).
Step 5. In part (b) since $f(x)$ and $g(x)$ are identical, we have $f'''(0)=g'''(0)$, and $g^{(4)}(x)= f^{(4)}(0)$
