Symmetry in the unit circle

Trigonometry: Angles with sine, cosine and tangent

Symmetry in the unit circle

We have already seen that the sine and cosine repeat themselves every $2 \pi$ . Now we will look at symmetry in the unit circle.

Symmetry

We can find three kinds of symmetry in the unit circle, namely reflection across the $x$ -axis, reflection across the $y$ -axis and reflection across the line $y=x$ . This symmetry gives us the following rules for sine and cosine.

Reflection across the	Sine	Cosine
$x$ -axis	$\sin(-\alpha) = -\sin(\alpha)$	$\cos(-\alpha) = \cos(\alpha)$
$y$ -axis	$\sin(\pi-\alpha) = \sin(\alpha)$	$\cos(\pi-\alpha) = -\cos(\alpha)$
line $y=x$	$\sin(\frac{\pi}{2}-\alpha) = \cos(\alpha)$	$\cos(\frac{\pi}{2}-\alpha) = \sin(\alpha)$

Through these rules, we only need to know the values of the sine and cosine in the first quarter of the unit circle, i.e. the values of sine and cosine for $0 \leq \alpha \leq \tfrac{\pi}{4}$ . $0 \leq \alpha \leq \tfrac{\pi}{2}$ .

Examples

Reflections cross the	Sine	Cosine	geogebra
$x$ -axis	$\begin{array}{rcl}\sin(\frac{5 \pi}{3})&=&\sin(-\frac{\pi}{3})\\ &=&-\sin(\frac{\pi}{3})\end{array}$	$\begin{array}{rcl}\cos(\frac{5 \pi}{3})&=&\cos(-\frac{\pi}{3})\\&=&\cos(\frac{\pi}{3})\end{array}$
$y$ -axis	$\begin{array}{rcl}\sin(\frac{2 \pi}{3})&=&\sin(\pi-\frac{\pi}{3})\\&=&\sin(\frac{\pi}{3})\end{array}$	$\begin{array}{rcl}\cos(\frac{2 \pi}{3})&=&\cos(\pi-\frac{\pi}{3})\\&=&-\cos(\frac{\pi}{3})\end{array}$
line $y=x$	$\begin{array}{rcl}\sin(\frac{1}{6}\pi)&=&\sin(\frac{\pi}{2}-\frac{\pi}{3})\\&=&\cos(\frac{\pi}{3})\end{array}$	$\begin{array}{rcl}\cos(\frac{1}{6}\pi)&=&\cos(\frac{\pi}{2}-\frac{\pi}{3})\\&=&\sin(\frac{\pi}{3})\end{array}$

Degrees For angles $\alpha$ in degrees the same rules apply, only the parts with $\pi$ are then converted to corresponding angles in degrees.

Reflection across the	Sine	Cosine
$x$ -axis	$\sin(360^\circ-\alpha) = -\sin(\alpha)$	$\cos(360^\circ-\alpha) = \cos(\alpha)$
$y$ -axis	$\sin(180^\circ-\alpha) = \sin(\alpha)$	$\cos(180^\circ-\alpha) = -\cos(\alpha)$
line $y=x$	$\sin(90^\circ-\alpha) = \cos(\alpha)$	$\cos(90^\circ-\alpha) = \sin(\alpha)$

If you know the values of the cosine and sine of the special angles between $0$ and $\frac{\pi}{2}$ , then you can calculate the values of special angles between $\frac{\pi}{2}$ and $2 \pi$ by using mirror symmetry.

Given is $\cos\left(\frac{\pi}{4}\right)=\dfrac{1}{\sqrt{2}}$ . What is $\cos\left(\frac{7 \pi}{4}\right)$ ?

$\cos\left(\frac{7 \pi}{4}\right)=\dfrac{1}{\sqrt{2}}$

The given angles are related to each other through reflection across the $x$ -axis. Hence $\cos\left(\frac{7 \pi}{4}\right)=\cos\left(\frac{\pi}{4}\right)=\dfrac{1}{\sqrt{2}}$

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