Fundamental identity with complementary arcs

Statement

Verify the identity $\sin^2\!\left(\tfrac{3\pi}{8}\right) + \cos^2\!\left(\tfrac{3\pi}{8}\right) = 1$ using the properties of associated arcs.

Solution

The fundamental identity holds for every angle, but here we are asked to verify it by passing through complementary arcs. Note that $\tfrac{3\pi}{8} = \tfrac{\pi}{2} - \tfrac{\pi}{8}$, so $\sin\!\left(\tfrac{3\pi}{8}\right) = \sin\!\left(\tfrac{\pi}{2}-\tfrac{\pi}{8}\right) = \cos\!\left(\tfrac{\pi}{8}\right),\qquad \cos\!\left(\tfrac{3\pi}{8}\right) = \cos\!\left(\tfrac{\pi}{2}-\tfrac{\pi}{8}\right) = \sin\!\left(\tfrac{\pi}{8}\right).$ Substituting: $\sin^2\!\left(\tfrac{3\pi}{8}\right) + \cos^2\!\left(\tfrac{3\pi}{8}\right) = \cos^2\!\left(\tfrac{\pi}{8}\right) + \sin^2\!\left(\tfrac{\pi}{8}\right) = \boxed{1},$ again by the fundamental identity applied to the angle $\tfrac{\pi}{8}$.

Links

Topics: Goniometry
Concepts: Associated arcs · Fundamental identity
Functions: Cosine · Sine
Skills: Proving · Using formulae
Exercise types: Proof