MathDB
Trigonometric Equality - IMO LongList 1979 - P6

Source:

May 29, 2011
trigonometrygeometrysimilar triangles

Problem Statement

Prove that 12⋅4sin⁡236∘−1=cos⁡72∘\frac 12 \cdot \sqrt{4\sin^2 36^{\circ} - 1}=\cos 72^\circ.
Back to ProblemsView on AoPS