This looks better $(4\cos^2 9^{\circ}-3) (4 \cos^2 27^{\circ}-3) = \tan 9^{\circ}$.

it helps me typing quicker, that's why i choose that

@below ... saw it Also a nice collection made by you for those PRMO 2016 papers as I had only the delhi region paper

it comes from Chandigarh Region, look here PS 2016 India PRMO post collection

actually parmenides you understood it wrongly these problems are from rmo chandigarh region 2016

Oh yes ... they are not from PRMO ... they are actually from RMO

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

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

I know but once u open the pdf it's written rmo

anyway, I renamed all the 8 problems, from preRMO to Chandigarh RMO

(soon they shall be added to the 2016 Indian RMO post collection)

parmenides51 wrote:

These are PRE RMO (Chandigarh region) problems It was the last year for which pre-RMO was a subjective pattern exam

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°