Symmetry properties and the Pythagorean identity
Explore the symmetry properties of the trigonometric ratios and the Pythagorean identity for Year 11 Specialist Mathematics in Queensland (QCAA). Reflecting a point on the unit circle connects the ratios of related angles, and one identity ties sine and cosine together at every angle.
You will learn the negative-angle, supplementary and complementary relations, use sin squared plus cos squared equals one to find exact values, and choose the correct sign from each quadrant — the identity toolkit behind exact-value work and later trigonometric proofs.
Theory
The symmetry properties of the trigonometric ratios and the Pythagorean identity are core tools in Year 11 Specialist Mathematics (QCAA, Queensland). Reflecting a point on the unit circle relates the ratios of angles such as \(-x\), \(180^\circ-x\) and \(90^\circ-x\), while \(\sin^2 x+\cos^2 x=1\) links sine and cosine at one angle. This page shows each relation, when to use it, and full worked examples.
Every angle \(x\) corresponds to a point \(P=(\cos x,\ \sin x)\) on the unit circle (the circle of radius \(1\) centred at the origin). Because \(P\) satisfies \(x^2+y^2=1\), the coordinates obey the Pythagorean identity \(\sin^2 x+\cos^2 x=1\).
Reflecting \(P\) in the \(x\)-axis sends the angle \(x\) to \(-x\) and keeps the \(x\)-coordinate but negates the \(y\)-coordinate. This gives the negative-angle relations: cosine is an even function, so \(\cos(-x)=\cos x\), while sine and tangent are odd functions, so \(\sin(-x)=-\sin x\) and \(\tan(-x)=-\tan x\).
Reflecting \(P\) in the \(y\)-axis sends \(x\) to the supplementary angle \(180^\circ-x\): \(\sin(180^\circ-x)=\sin x\), \(\cos(180^\circ-x)=-\cos x\) and \(\tan(180^\circ-x)=-\tan x\). Swapping the coordinates gives the complementary relations \(\sin(90^\circ-x)=\cos x\) and \(\cos(90^\circ-x)=\sin x\), where the co-functions trade places.
Together these let you rewrite the ratio of any angle in terms of a first-quadrant angle, so an exact value in one quadrant unlocks the same value (with the correct sign) everywhere else.
The Pythagorean identity for any angle \(x\):
Negative-angle (even/odd) relations:
Supplementary relations (angle \(180^\circ-x\)):
Complementary relations (angle \(90^\circ-x\)):
Finding an exact value using symmetry
- Locate the angle's quadrant, and note the sign of each ratio there (use CAST).
- Rewrite the angle as \(-x\), \(180^\circ-x\), \(90^\circ-x\) or a multiple, and apply the matching symmetry relation to bring it into the first quadrant.
- Read off the exact first-quadrant value (for \(30^\circ,45^\circ,60^\circ\), keep surds and fractions exact).
- Restore the sign from step 1; for an unknown ratio use \(\sin^2 x+\cos^2 x=1\) and pick the root that matches the quadrant.
Sine is an odd function, so reflecting in the \(x\)-axis negates the value:
| \(\sin(-30^\circ)\) | \(=\) | \(-\sin 30^\circ\) |
| \(\sin 30^\circ\) | \(=\) | \(\dfrac{1}{2}\) |
| \(\sin(-30^\circ)\) | \(=\) | \(-\dfrac{1}{2}\) |
\(\sin(-30^\circ)=-\dfrac{1}{2}\).
Write \(120^\circ=180^\circ-60^\circ\); cosine changes sign under this reflection:
| \(\cos 120^\circ\) | \(=\) | \(\cos(180^\circ-60^\circ)\) |
| \(=\) | \(-\cos 60^\circ\) | |
| \(\cos 60^\circ\) | \(=\) | \(\dfrac{1}{2}\) |
| \(\cos 120^\circ\) | \(=\) | \(-\dfrac{1}{2}\) |
\(\cos 120^\circ=-\dfrac{1}{2}\).
Rearrange the Pythagorean identity to find \(\sin^2 x\):
| \(\sin^2 x\) | \(=\) | \(1-\cos^2 x\) |
| \(=\) | \(1-\dfrac{25}{169}\) | |
| \(=\) | \(\dfrac{144}{169}\) |
In the third quadrant sine is negative, so take the negative root:
| \(\sin x\) | \(=\) | \(-\dfrac{12}{13}\) |
Divide to get the tangent:
| \(\tan x\) | \(=\) | \(\dfrac{\sin x}{\cos x}\) |
| \(=\) | \(\dfrac{-12/13}{-5/13}\) | |
| \(=\) | \(\dfrac{12}{5}\) |
\(\sin x=-\dfrac{12}{13}\) and \(\tan x=\dfrac{12}{5}\).
Replace \(1-\sin^2\theta\) using the Pythagorean identity, then cancel:
| \(\sin^2\theta+\cos^2\theta\) | \(=\) | \(1\) |
| \(1-\sin^2\theta\) | \(=\) | \(\cos^2\theta\) |
| \(\dfrac{1-\sin^2\theta}{\cos\theta}\) | \(=\) | \(\dfrac{\cos^2\theta}{\cos\theta}\) |
| \(=\) | \(\cos\theta\) |
\(\dfrac{1-\sin^2\theta}{\cos\theta}=\cos\theta\).
Common pitfalls
Frequently asked questions
Is sine an odd or even function?
Sine is odd, so \(\sin(-x)=-\sin x\). Tangent is also odd. Cosine is even, so \(\cos(-x)=\cos x\).
What is the Pythagorean identity?
For any angle \(x\), \(\sin^2 x+\cos^2 x=1\). It comes from the point \((\cos x,\sin x)\) lying on the unit circle \(x^2+y^2=1\).
How do I know whether cosine is positive or negative?
Use the quadrant. Cosine is positive in the first and fourth quadrants and negative in the second and third. The CAST diagram summarises which ratios are positive where.
What is the difference between supplementary and complementary angles?
Supplementary angles add to \(180^\circ\) and give relations like \(\sin(180^\circ-x)=\sin x\); complementary angles add to \(90^\circ\) and swap co-functions, \(\sin(90^\circ-x)=\cos x\).
How do I find an exact value like \(\sin 120^\circ\)?
Write \(120^\circ=180^\circ-60^\circ\) and use the supplementary relation: \(\sin 120^\circ=\sin 60^\circ=\dfrac{\sqrt{3}}{2}\).