Prove that $(4\cos^29^o – 3) (4 \cos^227^o– 3) = \tan 9^o$.
Problem
Source:
Tags: trigonometry
09.08.2019 14:54
This looks better $(4\cos^2 9^{\circ}-3) (4 \cos^2 27^{\circ}-3) = \tan 9^{\circ}$.
09.08.2019 14:56
it helps me typing quicker, that's why i choose that
09.08.2019 15:32
@below ... saw it Also a nice collection made by you for those PRMO 2016 papers as I had only the delhi region paper
09.08.2019 15:36
it comes from Chandigarh Region, look here PS 2016 India PRMO post collection
09.08.2019 16:03
09.08.2019 16:51
actually parmenides you understood it wrongly these problems are from rmo chandigarh region 2016
09.08.2019 16:59
Oh yes ... they are not from PRMO ... they are actually from RMO
09.08.2019 17:04
amaanmathbuddy_2006 wrote: actually parmenides you understood it wrongly these problems are from rmo chandigarh region 2016 was this any different pls dont double post
09.08.2019 17:25
amaanmathbuddy_2006 wrote: actually parmenides you understood it wrongly these problems are from rmo chandigarh region 2016 I shall correct as I return home, here it was named as pre-rmo 2016 from that region: https://www.pioneermathematics.com/maths-olympiad-rmo-inmo-imo.html
09.08.2019 19:31
I know but once u open the pdf it's written rmo
09.08.2019 22:33
anyway, I renamed all the 8 problems, from preRMO to Chandigarh RMO
(soon they shall be added to the 2016 Indian RMO post collection)
10.08.2019 17:03
parmenides51 wrote:
These are PRE RMO (Chandigarh region) problems It was the last year for which pre-RMO was a subjective pattern exam
08.01.2025 09:24
cos 3x = 3cosx - 4cos³x cos 3x/cos x = 3 - 4cos²x 4cos²9° - 3 = cos 27°/cos 9° —————(1) 4cos²27° - 3 = cos 81°/cos 27° —————(2) On multiplying equation (1) and (2), we get (4cos²9° - 3)(4cos²27° - 3) = tan 9°