Answer
See explanations.
Work Step by Step
(a) Step 1. Recall the formulas on page 818 as:
$A'=Acos^2\phi+Bsin\phi cos\phi +C sin^2\phi$, $B'=2(C-A)sin\phi cos\phi + B(cos^2\phi - sin^2\phi)$, and $C'=A sin^2\phi - B sin\phi cos\phi +C cos^2\phi$
Step 2. Evaluate $B'^2-4A'C'=(2(C-A)sin\phi cos\phi + B(cos^2\phi - sin^2\phi))^2-4(Acos^2\phi+Bsin\phi cos\phi +C sin^2\phi)(A sin^2\phi - B sin\phi cos\phi +C cos^2\phi)=4(C-A)^2sin^2\phi cos^2\phi +4B(C-A)sin\phi cos\phi (cos^2\phi - sin^2\phi) +B^2 (cos^2\phi - sin^2\phi)^2-4A^2sin^2\phi cos^2\phi +4ABsin\phi cos^3\phi -4ACcos^4\phi -4ABsin^3\phi cos\phi +4B^2sin^2\phi cos^2\phi -4BCsin\phi cos^3\phi -4ACsin^4\phi + 4BCsin^3\phi cos\phi - 4C^2sin^2\phi cos^2\phi$
Step 3. Combine the coefficients of like terms for $sin^4\phi, sin^3\phi cos\phi, sin^2\phi cos^2\phi, sin\phi cos^3\phi, cos^4\phi$ and use fundamental Identities such as $sin^2\phi + cos^2\phi =1$, we can reach our final conclusion that $B'^2-4A'C'=B^2-4AC$
(b) Evaluate $A'+C'=Acos^2\phi+Bsin\phi cos\phi +C sin^2\phi+A sin^2\phi - B sin\phi cos\phi +C cos^2\phi=A+C$, thus we proved that A + C is invariant under rotation.
(c) Recall that $F'=F$ from the formula on page 818, we can conclude that quantity F is invariant under rotation.
