Chapter A5.3, Problem 46E

(a)

To determine

To show: the numbers $A+C$ is invariant during rotation.

Answer to Problem 46E

Here, $A+C$ is invariant during rotation.

Explanation of Solution

Given: $A\text{'}+C\text{'}=A+CandD{\text{'}}^2+E{\text{'}}^2=D^2+E^2$.

Calculation:

It is known that

  $\begin{array}{l}A\prime =A{\cos }^2\alpha +B\cos \alpha \sin \alpha +C{\sin }^2\alpha \\ C\prime =A{\sin }^2\alpha +B\cos \alpha \sin \alpha +C{\cos }^2\alpha \end{array}$

Therefore,

  $\begin{array}{l}A\prime +C\prime =\left(A{\cos }^2\alpha +B\cos \alpha \sin \alpha +C{\sin }^2\alpha \right)+\left(A{\sin }^2\alpha +B\cos \alpha \sin \alpha +C{\cos }^2\alpha \right)\\ =A\left({\cos }^2\alpha +{\sin }^2\alpha \right)+C\left({\sin }^2\alpha +{\cos }^2\alpha \right)\\ =A+C\end{array}$

Hence, proved that $A+C$ is invariant during rotation.

Conclusion:

Hence, $A+C$ is invariant during rotation.

(b)

To determine

To show: the numbers $D^2+E^2$ is invariant during rotation.

Answer to Problem 46E

Here, $D^2+E^2$ is invariant during rotation.

Explanation of Solution

Calculation:

It is known that

  $\begin{array}{l}D\prime =D\cos \alpha +E\sin \alpha \\ E\prime =-D\sin \alpha +E\cos \alpha \end{array}$

Therefore,

  $\begin{array}{l}D{\prime }^2+E{\prime }^2={\left(D\cos \alpha +E\sin \alpha \right)}^2+{\left(-D\sin \alpha +E\cos \alpha \right)}^2\\ =D^2{\cos }^2\alpha +2DE\cos \alpha \sin \alpha +E^2{\sin }^2\alpha +D^2{\sin }^2\alpha -2DE\cos \alpha \sin \alpha +E^2{\cos }^2\alpha \\ =D^2\left({\cos }^2\alpha +{\sin }^2\alpha \right)+E^2\left({\sin }^2\alpha +{\cos }^2\alpha \right)\\ =D^2+E^2\end{array}$

Hence, proved that $D^2+E^2$ is invariant during rotation.

Conclusion:

Hence, $D^2+E^2$ is invariant during rotation.