MathDB

7

Part of VMEO I 2004

Problems(1)

product (4 sin^2 (2^{k}-1)\pi/2^{2005}} -3)

Source: 2004 VMEO I p7 Vietnamese Mathematics e - Olympiad https://artofproblemsolving.com/community/c2461015_vmeo__vietnam_mathematical

9/26/2021
Calculate the following P=(4sin⁡20−3)(4sin⁡2π22005−3)(4sin⁡22π22005−3)(4sin⁡23π22005−3)...P=(4\sin^2{0} -3)(4\sin^2\frac{\pi}{2^{2005}} -3)(4\sin^2\frac{2\pi}{2^{2005}} -3)(4\sin^2\frac{3\pi}{2^{2005}} -3)... ...    (4sin⁡2(22004−1)π22005−3)(4sin⁡2π2−3)...\,\,\,\,(4\sin^2\frac{(2^{2004}-1)\pi}{2^{2005}} -3)(4\sin^2\frac{\pi}{2} -3)
trigonometryProductalgebra