Symmetry Properties Of Trigonometric Functions
Understand the symmetry properties of sine and cosine for Queensland Year 11 Mathematical Methods (QCAA). Cosine is an even function and sine is an odd function, and related-angle rules connect any angle to a first-quadrant one.
You will learn to apply related-angle rules, use periodicity to simplify large angles, and combine reference angles with ASTC to find exact values anywhere on the circle in the QCAA course.
Every question with a fully worked solution.
- Symmetry Properties Of Trigonometric Functions - Video - Symmetry properties of circular functions Watch
Theory
In Year 11 Mathematical Methods (QCAA, Unit 1), the symmetry of the unit circle links the value of a trig function at one angle to its value at a related angle. Cosine is even and sine is odd, and the related-angle rules (such as \(\sin(\pi-t)=\sin t\)) let you evaluate any angle from a first-quadrant one. This page shows how to use these symmetries and periodicity.
Reflecting a point on the unit circle produces a symmetry. Reflecting across the \(x\)-axis sends \(t\) to \(-t\): the \(x\)-coordinate is unchanged and the \(y\)-coordinate reverses. So cosine is even, \(\cos(-t)=\cos t\), and sine is odd, \(\sin(-t)=-\sin t\).
Reflecting across the \(y\)-axis sends \(t\) to \(\pi-t\), giving the related-angle rules \(\sin(\pi-t)=\sin t\) and \(\cos(\pi-t)=-\cos t\). A half-turn to \(\pi+t\) gives \(\sin(\pi+t)=-\sin t\) and \(\cos(\pi+t)=-\cos t\); and \(\sin(2\pi-t)=-\sin t\), \(\cos(2\pi-t)=\cos t\).
Because the circle repeats every turn, the functions are periodic: \(\sin(t+2\pi)=\sin t\) and \(\cos(t+2\pi)=\cos t\). Reduce a large angle by whole turns first, then use symmetry.
Even/odd symmetry:
Related-angle rules from reflections and half-turns:
Periodicity (repeat every full turn):
How to evaluate using symmetry
- Reduce: subtract whole turns \(2\pi\) so the angle lies in \([0,\,2\pi)\).
- Rewrite the angle as \(\pi-t\), \(\pi+t\), \(2\pi-t\) or \(-t\) about a first-quadrant reference angle.
- Apply the rule: take the first-quadrant value and attach the ASTC sign for the quadrant.
Cosine is even, so the value is unchanged:
| \(\cos(-t)\) | \(=\) | \(\cos t\) |
| \(=\) | \(\dfrac{3}{5}\) |
Sine is odd, so the value reverses sign:
| \(\sin(-t)\) | \(=\) | \(-\sin t\) |
| \(=\) | \(-\dfrac{4}{5}\) |
\(\cos(-t)=\dfrac{3}{5}\) and \(\sin(-t)=-\dfrac{4}{5}\).
Write \(\dfrac{5\pi}{6}=\pi-\dfrac{\pi}{6}\) and use \(\sin(\pi-t)=\sin t\):
| \(\sin\dfrac{5\pi}{6}\) | \(=\) | \(\sin\!\left(\pi-\dfrac{\pi}{6}\right)\) |
| \(=\) | \(\sin\dfrac{\pi}{6}\) |
Read the first-quadrant value:
| \(=\) | \(\dfrac{1}{2}\) |
\(\sin\dfrac{5\pi}{6}=\dfrac{1}{2}\).
Write \(\dfrac{7\pi}{6}=\pi+\dfrac{\pi}{6}\) and use \(\cos(\pi+t)=-\cos t\):
| \(\cos\dfrac{7\pi}{6}\) | \(=\) | \(\cos\!\left(\pi+\dfrac{\pi}{6}\right)\) |
| \(=\) | \(-\cos\dfrac{\pi}{6}\) |
Substitute the first-quadrant value:
| \(=\) | \(-\dfrac{\sqrt3}{2}\) |
\(\cos\dfrac{7\pi}{6}=-\dfrac{\sqrt3}{2}\).
The angle is more than \(2\pi\); subtract one whole turn \(2\pi=\dfrac{12\pi}{6}\):
| \(\sin\dfrac{13\pi}{6}\) | \(=\) | \(\sin\!\left(\dfrac{13\pi}{6}-2\pi\right)\) |
| \(=\) | \(\sin\dfrac{\pi}{6}\) |
Read the first-quadrant value:
| \(=\) | \(\dfrac{1}{2}\) |
\(\sin\dfrac{13\pi}{6}=\dfrac{1}{2}\).
Common pitfalls
Frequently asked questions
Is cosine even or odd?
Cosine is even: \(\cos(-t)=\cos t\), because reflecting in the \(x\)-axis leaves the \(x\)-coordinate unchanged. Sine is odd: \(\sin(-t)=-\sin t\).
What is the rule for sin(pi minus t)?
\(\sin(\pi-t)=\sin t\). The angles \(t\) and \(\pi-t\) are reflections in the \(y\)-axis, so they have the same \(y\)-coordinate and the same sine.
How do you evaluate sin or cos of an angle bigger than 2 pi?
Use periodicity: subtract whole turns of \(2\pi\) until the angle lies in \([0,\,2\pi)\), then evaluate. For example \(\sin\dfrac{13\pi}{6}=\sin\dfrac{\pi}{6}=\dfrac{1}{2}\).
Why is cos(pi + t) equal to negative cos t?
Adding \(\pi\) moves to the diametrically opposite point, reversing both coordinates, so \(\cos(\pi+t)=-\cos t\) and \(\sin(\pi+t)=-\sin t\).
How do the related-angle rules connect to ASTC?
Each rule gives the first-quadrant size with a sign; that sign is exactly what ASTC predicts for the quadrant the angle lands in.