Elementary inequalities with sine and cosine

The simplest trigonometric inequalities are read directly off the trigonometric circle, with no calculation needed: it is enough to identify the arcs corresponding to the points that satisfy the condition.

In summary — Elementary inequalities

They are read directly off the trigonometric circle.

• $\sin x \ge k$ (with $|k|\le 1$): these are the angles corresponding to the points of the circle above the horizontal line $y=k$. The solution, on the interval $[0,2\pi)$, is an arc; it must then be generalised by adding $2k_1\pi$.
• $\cos x \ge k$ (with $|k|\le 1$): the points to the right of the vertical line $x=k$.

For an inequality with the sine one draws the horizontal line $y=k$: the condition $\sin x \ge k$ is satisfied by the points of the circle lying above that line.

The inequality $\sin x \ge 1/2$, on the interval $[0,2\pi)$, has as its solution the green arc above the line $y=1/2$, that is $x\in\left[\dfrac{\pi}{6}, \dfrac{5\pi}{6}\right]$.

For the cosine the reasoning is entirely analogous, but with a vertical line $x=k$: the condition $\cos x \ge k$ holds for the points to the right of the line, the condition $\cos x \le k$ for the points to the left.

Example

Solve $\cos x < -\dfrac{\sqrt{2}}{2}$ on $[0,2\pi)$.

The values of $x$ for which $\cos x = -\dfrac{\sqrt{2}}{2}$ are $\dfrac{3\pi}{4}$ and $\dfrac{5\pi}{4}$. The inequality $\cos x < -\dfrac{\sqrt{2}}{2}$ requires the points of the trigonometric circle to the left of the vertical line $x=-\dfrac{\sqrt{2}}{2}$: it is the arc from $\dfrac{3\pi}{4}$ to $\dfrac{5\pi}{4}$ (endpoints excluded). $x\in\left(\frac{3\pi}{4} + 2k\pi,\ \frac{5\pi}{4} + 2k\pi\right)$

Links

Topics: Trigonometric inequalities
Concepts: Trigonometric circle · Trigonometric inequality · Reading a trigonometric inequality on the circle
Functions: Cosine · Sine
Methods: Trigonometric inequalities
Skills: Interpreting a graph · Solving inequalities