Is $\sin\left[2\alpha\right]\cos\left[2\alpha\right]\ge0$ a valid restriction on the angles of the principal stresses in 2D elasticity? This question pertains to Elasticity: Tensor, Dyadic, and Engineering Approaches By: Pei Chi Chou, Nicholas J. Pagano, Section 1.4.
The objective under discussion is to find the directions of stationary normal stress. The following transformation equations have been established:
$$\sigma^{\overline{x}}=\sigma^{x}\cos^{2}\left[\alpha\right]+\sigma^{y}\sin^{2}\left[\alpha\right]+2\tau^{x}{}_{y}\cos\left[\alpha\right]\sin\left[\alpha\right],\tag{1.5}$$
$$\sigma^{\overline{y}}=\sigma^{x}\sin^{2}\left[\alpha\right]+\sigma^{y}\cos^{2}\left[\alpha\right]-2\tau^{x}{}_{y}\cos\left[\alpha\right]\sin\left[\alpha\right].\tag{1.7}$$
Using the trigonometric identities
$$\sin\left[2\alpha\right]=2\cos\left[\alpha\right]\sin\left[\alpha\right],\tag{1.8a}$$
$$\sin^{2}\left[\alpha\right]=\frac{1}{2}\left(1-\cos\left[2\alpha\right]\right),\tag{1.8b}$$
$$\cos^{2}\left[\alpha\right]=\frac{1}{2}\left(1+\cos\left[2\alpha\right]\right),\tag{1.8c}$$
these become
$$\sigma^{\overline{x}}=\frac{\sigma^{x}+\sigma^{y}}{2}+\left(\frac{\sigma^{x}-\sigma^{y}}{2}\cos\left[2\alpha\right]+\tau^{x}{}_{y}\sin\left[2\alpha\right]\right),\tag{1.9a}$$
$$\sigma^{\overline{y}}=\frac{\sigma^{x}+\sigma^{y}}{2}-\left(\frac{\sigma^{x}-\sigma^{y}}{2}\cos\left[2\alpha\right]+\tau^{x}{}_{y}\sin\left[2\alpha\right]\right).\tag{1.9b}$$
We set the derivative with respect to $\alpha$ of either of these equations ($\sigma^{\overline{x}}$ in this case) equal to zero 
$$\left(\sigma^{x}-\sigma^{y}\right)\sin\left[2\alpha\right]=2\tau^{x}{}_{y}\cos\left[2\alpha\right],\tag{1.11}$$
and find the two roots $\left\{ 2\alpha_{1},2\alpha_{2}\right\}$  of the resulting trigonometric expression 
$$\tan\left[2\alpha\right]=\frac{2\tau^{x}{}_{y}}{\sigma^{x}-\sigma^{y}}.\tag{1.12}$$
It follows that $2\alpha_{2}=2\alpha_{1}\pm\pi;$ thus $\alpha_{2}=\alpha_{1}\pm\frac{\pi}{2}.$ The sine and cosine of $2\alpha$ are
$$\sin\left[2\alpha\right]=\pm\frac{2\tau^{x}{}_{y}}{\sqrt{4\left(\tau^{x}{}_{y}\right)^{2}+\left(\sigma^{x}-\sigma^{y}\right)^{2}}},\tag{1.12a}$$
$$\cos\left[2\alpha\right]=\pm\frac{\sigma^{x}-\sigma^{y}}{\sqrt{4\left(\tau^{x}{}_{y}\right)^{2}+\left(\sigma^{x}-\sigma^{y}\right)^{2}}}.\tag{1.12b}$$
In the first sentence of page 10, the book claims that $\sin\left[2\alpha\right]$ and $\cos\left[2\alpha\right]$ are either both positive or both negative. That would restrict $\alpha$ to $0\le\alpha\le\frac{\pi}{4}$ or $\frac{\pi}{2}\le\alpha\le\frac{3\pi}{4}$. More significantly, it would require $\tau^{x}{}_{y}$ and $\sigma^{x}-\sigma^{y}$ to have the same arithmetic sign. I see no justification for either of those restrictions. Is the claim that $\sin\left[2\alpha\right]\cos\left[2\alpha\right]\ge0$ valid?
The assertion that arithmetic signs of $\sin\left[2\alpha\right]$ and $\cos\left[2\alpha\right]$ must match appears to contradict the discussion in the final paragraph on page 10 in which the case of 
$$\frac{\pi}{2}<2\alpha\iff2\tau^{x}{}_{y}>0\wedge\left(\sigma^{x}-\sigma^{y}\right)<0$$ 
is considered. That clearly makes the signs of eq 1.12a and eq 1.12b different.
It might be the case that any system can be fully characterized by considering a range of angles which conforms to $\sin\left[2\alpha\right]\cos\left[2\alpha\right]\ge0$ , but that is not the claim made by the authors.
 A: Ref. 1 writes [admittedly somewhat confusingly]:

[...], and noting that the sine and cosine are either both plus or both minus, [...]

Ref. 1 does not say that sine and cosine are either both positive or either both negative in eqs. (1.12a)-(1.12b), which would have been incorrect$^{\dagger}$. Rather Ref. 1 is trying to say [that it follows from eq. (1.12)] that either both the upper $+$ signs in the $\pm$ symbol apply, or both the lower $-$ signs in the $\pm$ symbol apply, but a mixture with one upper $+$ sign and one lower $-$ sign is not allowed.
References: 


*

*Pei Chi Chou & Nicholas J. Pagano, Elasticity: Tensor, Dyadic, and Engineering Approaches, 1967; p.9-10.


$^{\dagger}$ The variable $\alpha$ is $2\pi$-periodic.
A: The statement in the book has to be viewed through the lens of a particular method of proof.  My original effort was not consistent with that approach.  I now believe I can state the intent of the authors in a more satisfying manner.
Define the vector constant
$$\mathfrak{b}\equiv\begin{bmatrix}\frac{1}{2}\left(\sigma^{x}-\sigma^{y}\right)\\
\tau^{x}{}_{y}
\end{bmatrix}\equiv b\begin{bmatrix}\cos\left[\beta\right]\\
\sin\left[\beta\right]
\end{bmatrix}\equiv b\hat{\mathfrak{r}}\left[\beta\right],$$
the vector variable
$$\hat{\mathfrak{r}}\left[2\alpha\right]\equiv\begin{bmatrix}\cos\left[2\alpha\right]\\
\sin\left[2\alpha\right]
\end{bmatrix},$$
and the scalar the constant $\left\langle \sigma\right\rangle \equiv\frac{\sigma^{x}+\sigma^{y}}{2}.$ The first normal stress transformation equation (1.9a) is now written
$$\sigma^{\overline{x}}=\left\langle \sigma\right\rangle +\mathfrak{b}\cdot\hat{\mathfrak{r}}\left[2\alpha\right].$$
Setting the derivative with respect to $\alpha$ equal to $0$ produces
$$0=\hat{\mathfrak{r}}\left[\beta\right]\cdot\hat{\mathfrak{r}}\left[2\alpha+\frac{\pi}{2}\right].$$
Which implies
$$\hat{\mathfrak{r}}\left[2\alpha\right]=\pm\hat{\mathfrak{r}}\left[\beta\right].$$
So that
$$b\hat{\mathfrak{r}}\left[2\alpha\right]=\pm\frac{1}{2}\begin{bmatrix}\sigma^{x}-\sigma^{y}\\
2\tau^{x}{}_{y}
\end{bmatrix}=\pm\begin{bmatrix}\cos\left[2\alpha\right]\\
\sin\left[2\alpha\right]
\end{bmatrix},\text{and}$$
$$\tan\left[2\alpha\right]=\frac{\pm2\tau^{x}{}_{y}}{\pm\left(\sigma^{x}-\sigma^{y}\right)}.$$
