Chapter: Trigonometry· Mathematics Part 2 (Geometry)
Answer: Proved using Pythagoras theorem.
[Figure: ΔABC with \(\angle ABC = 90^\circ\) and \(\angle ACB = \theta\).]
In ΔABC, by Pythagoras theorem, \(AB^2 + BC^2 = AC^2\).
Dividing both sides by \(AC^2\): \(\dfrac{AB^2}{AC^2} + \dfrac{BC^2}{AC^2} = 1\)
\(\therefore \left(\dfrac{AB}{AC}\right)^2 + \left(\dfrac{BC}{AC}\right)^2 = 1\)
But \(\dfrac{AB}{AC} = \sin\theta\) (opposite side / hypotenuse) and \(\dfrac{BC}{AC} = \cos\theta\) (adjacent side / hypotenuse).
\(\therefore \sin^2\theta + \cos^2\theta = 1\).