Further Symmetry Properties And The Pythagorean Identity
Learn the Pythagorean identity for Queensland Year 11 Mathematical Methods (QCAA). Sine squared plus cosine squared always equals one, and tangent equals sine divided by cosine — relations that tie the three ratios together.
You will learn to find a missing ratio from a known one and its quadrant, attach the correct sign using ASTC, and apply complementary and symmetry relations to simplify expressions — a powerful toolkit for exact-value work in the QCAA course.
Every question with a fully worked solution.
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Theory
In Year 11 Mathematical Methods (QCAA, Unit 1), the Pythagorean identity \(\sin^2 t+\cos^2 t=1\) links sine and cosine, and \(\tan t=\dfrac{\sin t}{\cos t}\). Given one ratio and the quadrant, you can find the others exactly, choosing the correct sign from ASTC. Symmetry relations such as \(\sin\!\left(\dfrac{\pi}{2}-t\right)=\cos t\) also follow from the unit circle.
The Pythagorean identity \(\sin^2 t+\cos^2 t=1\) holds for every angle \(t\); it is just Pythagoras applied to the right triangle inside the unit circle, whose legs are \(\cos t\) and \(\sin t\) and whose hypotenuse is \(1\).
The tangent ratio is \(\tan t=\dfrac{\sin t}{\cos t}\). So once you know \(\sin t\) and \(\cos t\) you can find \(\tan t\) exactly.
Rearranging the identity gives \(\sin t=\pm\sqrt{1-\cos^2 t}\) and \(\cos t=\pm\sqrt{1-\sin^2 t}\); the quadrant chooses the sign. Complementary relations like \(\sin\!\left(\dfrac{\pi}{2}-t\right)=\cos t\) and \(\cos\!\left(\dfrac{\pi}{2}-t\right)=\sin t\) come from the symmetry of the circle.
The core identities:
How to find a missing ratio
- Substitute the known ratio into \(\sin^2 t+\cos^2 t=1\).
- Solve for the square of the missing ratio, then square-root to get its size.
- Sign: use the given quadrant and ASTC to choose \(+\) or \(-\).
- Tangent: if needed, form \(\tan t=\dfrac{\sin t}{\cos t}\) from the two signed ratios.
Pythagoras — substitute \(\sin t=\dfrac35\):
| \(\cos^2 t\) | \(=\) | \(1-\sin^2 t\) |
| \(=\) | \(1-\left(\tfrac35\right)^2\) | |
| \(=\) | \(1-\tfrac{9}{25}\) | |
| \(=\) | \(\tfrac{16}{25}\) |
Sign — Q1, so cosine is positive:
| \(\cos t\) | \(=\) | \(+\sqrt{\tfrac{16}{25}}\) |
| \(=\) | \(\tfrac45\) |
Tangent — \(\dfrac{\sin t}{\cos t}\):
| \(\tan t\) | \(=\) | \(\dfrac{3/5}{4/5}\) |
| \(=\) | \(\tfrac34\) |
\(\cos t=\dfrac45,\ \ \tan t=\dfrac34\).
Pythagoras:
| \(\sin^2 t\) | \(=\) | \(1-\cos^2 t\) |
| \(=\) | \(1-\left(\tfrac{5}{13}\right)^2\) | |
| \(=\) | \(1-\tfrac{25}{169}\) | |
| \(=\) | \(\tfrac{144}{169}\) |
Sign — \(t\) is in Q2, so sine is positive:
| \(\sin t\) | \(=\) | \(+\sqrt{\tfrac{144}{169}}\) |
| \(=\) | \(\tfrac{12}{13}\) |
\(\sin t=\dfrac{12}{13}\).
Pythagoras:
| \(\cos^2 t\) | \(=\) | \(1-\sin^2 t\) |
| \(=\) | \(1-\left(\tfrac12\right)^2\) | |
| \(=\) | \(1-\tfrac14\) | |
| \(=\) | \(\tfrac34\) |
Sign — Q3, so cosine is negative:
| \(\cos t\) | \(=\) | \(-\sqrt{\tfrac34}\) |
| \(=\) | \(-\dfrac{\sqrt3}{2}\) |
\(\cos t=-\dfrac{\sqrt3}{2}\).
Rewrite the denominator with the identity \(1-\sin^2 t=\cos^2 t\):
| \(\dfrac{\sin t\cos t}{1-\sin^2 t}\) | \(=\) | \(\dfrac{\sin t\cos t}{\cos^2 t}\) |
Cancel one factor of \(\cos t\):
| \(=\) | \(\dfrac{\sin t}{\cos t}\) |
Recognise the tangent ratio:
| \(=\) | \(\tan t\) |
\(\dfrac{\sin t\cos t}{1-\sin^2 t}=\tan t\).
Common pitfalls
Frequently asked questions
What is the Pythagorean identity?
\(\sin^2 t+\cos^2 t=1\) for every angle \(t\). It comes from Pythagoras on the right triangle inside the unit circle.
How do I find cos t if I know sin t?
Rearrange to \(\cos t=\pm\sqrt{1-\sin^2 t}\), then choose the sign from the quadrant using ASTC.
How do I choose the plus or minus sign?
Use the quadrant: cosine is positive in Q1 and Q4, sine is positive in Q1 and Q2. The stated range for \(t\) tells you the quadrant.
How is tan t related to sin and cos?
\(\tan t=\dfrac{\sin t}{\cos t}\). Once you have both signed ratios, divide to get the tangent.
What is a complementary angle relation?
\(\sin\!\left(\dfrac{\pi}{2}-t\right)=\cos t\) and \(\cos\!\left(\dfrac{\pi}{2}-t\right)=\sin t\); the sine of an angle equals the cosine of its complement.