MathDB

6

Part of 1979 IMO Longlists

Problems(1)

Trigonometric Equality - IMO LongList 1979 - P6

Source:

5/29/2011
Prove that 12⋅4sin⁡236∘−1=cos⁡72∘\frac 12 \cdot \sqrt{4\sin^2 36^{\circ} - 1}=\cos 72^\circ.
trigonometrygeometrysimilar triangles