= Solution
Choose the following consistent sign convention. The <Givens rotation> $J(i,j,\theta)$ equals the <identity matrix> except for the block on coordinates $i,j$:
$$
J\big|_{i,j}=\begin{pmatrix}\cos\theta&\sin\theta\\-\sin\theta&\cos\theta\end{pmatrix}.
$$
Thus it changes only the $i$th and $j$th coordinates of a <vector>. Write $c=\cos\theta$, $s=\sin\theta$. Its two-coordinate block satisfies
$$
\begin{pmatrix}c&-s\\s&c\end{pmatrix}
\begin{pmatrix}c&s\\-s&c\end{pmatrix}
=\begin{pmatrix}c^2+s^2&0\\0&c^2+s^2\end{pmatrix}=I_2.
$$
All remaining coordinates are unchanged, so $J^TJ=I_n$. Hence \b[every such <Givens rotation> is an <orthogonal matrix>]. With this convention the two-coordinate <rotation matrix> rotates through angle $-\theta$.
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