y = \sin^2 x

Graph of the Function $y = \sin^2 x$

$y = \sin^2 x$ is the sine squared. Squaring folds the negative part upward and halves the period. It is the shape that recurs whenever the strength of an oscillation is at stake, from the root mean square of an alternating current to the intensity of light.

The half-angle formula

Solving the double-angle identity $\cos 2x = 1 - 2\sin^2 x$ gives the following.

\sin^2 x = \frac{1 - \cos 2x}{2}

As this shows, $\sin^2 x$ is nothing but $\cos 2x$ turned upside down, halved and lifted by $\dfrac{1}{2}$ : a cosine wave of amplitude $\dfrac{1}{2}$ about the level $\dfrac{1}{2}$ , with period $\pi$ .

Domain and range

The domain is all real numbers
The range is $0 \leq y \leq 1$
The period is $\pi$
It is an even function

Period and symmetry

Although $\sin(x+\pi) = -\sin x$ , squaring erases the sign, so the period is $\pi$ , half that of $\sin x$ itself. That halving is easy to overlook. Squaring an odd function also produces an even one, so the graph is symmetric about the $y$ -axis.

Zeros and maxima

The zeros are at $x = n\pi$ , where $\sin x = 0$ , but they are double zeros: the curve touches the $x$ -axis without passing below it. The maximum value $1$ occurs at $x = \dfrac{\pi}{2} + n\pi$ . Unlike $\sin x$ with its alternating crests and troughs, this graph is a row of evenly spaced crests only.

$x$	$\sin^2 x$	Position
$0$	$0$	touches the $x$ -axis
$\dfrac{\pi}{4}$	$\dfrac{1}{2}$	inflection point
$\dfrac{\pi}{2}$	$1$	maximum
$\dfrac{3\pi}{4}$	$\dfrac{1}{2}$	inflection point
$\pi$	$0$	touches the $x$ -axis

Monotonicity and inflection

The double-angle formulas keep the derivatives short.

\begin{align*} y' &= 2\sin x\cos x = \sin 2x \\ y'' &= 2\cos 2x \end{align*}

The inflection points are where $\cos 2x = 0$ , at $x = \dfrac{\pi}{4} + \dfrac{n\pi}{2}$ , and the value there is $\dfrac{1}{2}$ , exactly halfway up.

The average is one half

The identity $\sin^2 x + \cos^2 x = 1$ holds everywhere, and $\cos^2 x$ is merely a translate of $\sin^2 x$ , so the two must have the same average. Since their sum averages $1$ , each averages $\dfrac{1}{2}$ . Integration gives the same answer.

\frac{1}{\pi}\int_0^{\pi}\sin^2 x\,dx = \frac{1}{2}

The antiderivative is $\dfrac{x}{2} - \dfrac{\sin 2x}{4} + C$ .

Applications

Power dissipated by an alternating current is proportional to the square of the current, so a sinusoidal current produces power of exactly this shape. The average being $\dfrac{1}{2}$ is what makes the root mean square value equal to $\dfrac{1}{\sqrt{2}}$ times the peak, a cornerstone of electrical engineering¹.

The root mean square and the power of an alternating current
Malus's law, giving the intensity of light passed by a polarizer²
The probability density of quantum mechanics, the squared modulus of a wave function

Root mean square, Wikipedia
Polarizer, Wikipedia

Graph of the Function y=sin⁡2xy = \sin^2 xy=sin2x