how to know in which range $\cos^{2}\theta $ $\le$ $\sin^{2}\theta$? how to know in which range $\cos^{2}\theta $ $\le$ $\sin^{2}\theta$? I know this might be an easy question, but it has been so long since I dealt with trigonometric inequalities, for all I can do now is just to look at their graphs and determine by that, but is there an easier way?  
 A: Since $\sin^2\theta+\cos^2\theta=1$, your inequality holds iff $\sin^2\theta\geq\frac{1}{2}$, hence iff:
$$ \theta \in \left[\frac{\pi}{4},\frac{3\pi}{4}\right]+\pi\mathbb{Z}. $$
A: Rearranging gives $$\cos^2 \theta - \sin^2 \theta \leq 0 \\ \cos(2\theta) \leq 0$$
[edit]
Using the notation from Jack D'Aurizio's answer, we can now write $$2\theta \in \left[-\frac \pi 2, \frac \pi 2 \right] + 2\pi\Bbb Z$$ so $$\theta \in \left[ -\frac \pi 4, \frac \pi 4 \right] + \pi\Bbb Z$$
A: Hint: this is equivalent to $2cos^2(\theta) \leq 1$.
A: Geometrically, this is obvious: $(\cos \theta,\sin \theta)$ are the coordinates $(x,y)$ of a point of the unit circle. This inequality is equivalent to $\lvert x\rvert \le\lvert y\rvert $, and the solutions are the points of  the plane in the union of the quadrants determined by the straight lines $y=x,\enspace y=-x$ which contain the $y$-axis.The latter intersects the unit circle at $\theta= \pm\dfrac\pi4, \pm\dfrac{3\pi}4$. Whence the solutions:
$$\theta\in \bigcup_{k\in\mathbf Z}\Bigl(\Bigl[\dfrac\pi4+2k\pi,\dfrac{3\pi}4+2k\pi\Bigr]\cup\Bigl[-\dfrac{3\pi}4+2k\pi,-\dfrac\pi4+2k\pi\Bigr]\Bigr)$$
A: Use the fact that $\sin^2\theta+\cos^2\theta=1$ to rewrite the inequality as:
$$\sin\theta \geq \frac{\sqrt 2}{ 2}\qquad\text{or}\qquad \sin\theta \leq -\frac{\sqrt 2}{2}$$
that should be easier for you.
A: Dividing by $\cos^2\theta$ this is the same as finding when $\tan^2\theta\geq1$
and this happens for $\theta\geq\pi/4$. And more by periodicity.
EDIT: (Suggested by Timbuc's comment) AT points that are odd multiples of $\frac12\pi$ tan function has singularity; however $\cos^2\theta \leq \sin^2\theta$ is valid. I meant to first see in the open interval $-(\pi/2, pi/2)$ then periodical extended intervals.
