Question

In(6.14), let $\mathbf{\theta }\mathbf{=}\mathbf{\varphi }\mathbf{=}\mathbf{\tau \tau }\mathbf{/}\mathbf{2}$ and verify the result numerically.

Short answer

The relation $\left(\begin{array}{cc}\cos \varphi & -\sin \mathrm{\varphi }\\ \sin \mathrm{\varphi }& \cos \mathrm{\varphi }\end{array}\right)\left(\begin{array}{cc}\cos \mathrm{\theta }& -\sin \mathrm{\theta }\\ \sin \mathrm{\theta }& \cos \mathrm{\theta }\end{array}\right)=\left(\begin{array}{cc}\cos (\mathrm{\theta }+\mathrm{\varphi })& -\sin (\mathrm{\theta }+\mathrm{\varphi })\\ \sin (\mathrm{\theta }+\mathrm{\varphi })& \cos (\mathrm{\theta }+\mathrm{\varphi })\end{array}\right)$is verified numerically for and its numerical value is $\left(\begin{array}{cc}-1& 0\\ 0& -1\end{array}\right)$.

Step-by-step solution

The trigonometric values of sine and cosine function for different angles:

The numerical values of the sine and cosine function at $\frac{\mathbf{\pi }}{\mathbf{2}}$, and $\mathbf{\pi }$are

$$\begin{gathered} \mathit{s}\mathit{i}\mathit{n}\left(\frac{\pi }{2}\right)\mathbf{=}\mathbf{1} \\ \mathit{c}\mathit{o}\mathit{s}\left(\frac{\pi }{2}\right)\mathbf{=}\mathbf{0} \\ \mathit{s}\mathit{i}\mathit{n}\mathbf{}\left(\pi \right)\mathbf{=}\mathbf{0} \\ \mathit{c}\mathit{o}\mathit{s}\mathbf{}\left(\pi \right)\mathbf{=}\mathbf{-}\mathbf{1} \end{gathered}$$

Given parameters:

The result (6.14) is $\left(\begin{array}{cc}\cos \varphi & -\sin \varphi \\ \sin \varphi & \cos \mathrm{\varphi }\end{array}\right)\left(\begin{array}{cc}\cos \mathrm{\theta }& -\sin \mathrm{\theta }\\ \sin \mathrm{\theta }& \cos \mathrm{\theta }\end{array}\right)=\left(\begin{array}{cc}\cos (\mathrm{\theta }+\mathrm{\varphi })& -\sin (\mathrm{\theta }+\mathrm{\varphi })\\ \sin (\mathrm{\theta }+\mathrm{\varphi })& \cos (\mathrm{\theta }+\mathrm{\varphi })\end{array}\right)$.

It needs to be verified for $\mathrm{\theta }=\mathrm{\varphi }=\mathrm{\pi }/2$.

Finding the numerical values of the left-hand side and the right-hand side of the relation:

Substitute into the left-hand side of the relation $(6.14)$and evaluate the result.

$\left(\begin{array}{cc}\cos \frac{\mathrm{\pi }}{2}& -\sin \frac{\mathrm{\pi }}{2}\\ \sin \frac{\mathrm{\pi }}{2}& \cos \frac{\mathrm{\pi }}{2}\end{array}\right)\left(\begin{array}{cc}\cos \frac{\mathrm{\pi }}{2}& -\sin \frac{\mathrm{\pi }}{2}\\ \sin \frac{\mathrm{\pi }}{2}& \cos \frac{\mathrm{\pi }}{2}\end{array}\right)=\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)$

Multiply the matrices.

$\left(\begin{array}{cc}\cos \frac{\mathrm{\pi }}{2}& -\sin \frac{\mathrm{\pi }}{2}\\ \sin \frac{\mathrm{\pi }}{2}& \cos \frac{\mathrm{\pi }}{2}\end{array}\right)\left(\begin{array}{cc}\cos \frac{\mathrm{\pi }}{2}& -\sin \frac{\mathrm{\pi }}{2}\\ \sin \frac{\mathrm{\pi }}{2}& \cos \frac{\mathrm{\pi }}{2}\end{array}\right)=\left(\begin{array}{cc}-1& 0\\ 0& -1\end{array}\right)$

Substitute $\mathrm{\theta }=\mathrm{\varphi }=\mathrm{\pi }/2$into the right-hand side of the relation (6.14).

role="math" localid="1658989214239" $\left(\begin{array}{cc}\cos \left(\frac{\mathrm{\pi }}{2}+\frac{\mathrm{\pi }}{2}\right)& -\sin \left(\frac{\mathrm{\pi }}{2}+\frac{\mathrm{\pi }}{2}\right)\\ \sin \left(\frac{\mathrm{\pi }}{2}+\frac{\mathrm{\pi }}{2}\right)& \cos \left(\frac{\mathrm{\pi }}{2}+\frac{\mathrm{\pi }}{2}\right)\end{array}\right)=\left(\begin{array}{cc}\cos \left(\mathrm{\pi }\right)& -\sin \left(\mathrm{\pi }\right)\\ \sin \left(\mathrm{\pi }\right)& \cos \left(\mathrm{\pi }\right)\end{array}\right)$

Evaluate the sine and cosine function.

$\left(\begin{array}{cc}\cos \left(\mathrm{\pi }\right)& -\sin \left(\mathrm{\pi }\right)\\ \sin \left(\mathrm{\pi }\right)& \cos \left(\mathrm{\pi }\right)\end{array}\right)=\left(\begin{array}{cc}-1& 0\\ 0& -1\end{array}\right)$

Hence, the relation (6.14) is verified numerically