# Misc 10

Last updated at Dec. 8, 2016 by Teachoo

Last updated at Dec. 8, 2016 by Teachoo

Transcript

Misc, 10 Find sin π₯/2, cos π₯/2 and tan π₯/2 for sinβ‘π₯ = 1/4 , π₯ in quadrant II Given that x is in quadrant II So, 90Β° < x < 180Β° Replacing x with π₯/2 (90Β°)/2 < π₯/2 < (180Β°)/2 45Β° < π₯/2 < 90Β° So, π₯/2 lies in Ist quadrant In Ist quadrant, sin , cos & tan are positive β sin π₯/2 , cos π₯/2 and tan π₯/2 are positive Given sin x = 1/4 sin x = 2 sin π₯/2 cos π₯/2 1/4 = 2 sin π₯/2 cos π₯/2 2 sin π₯/2 cos π₯/2 = 1/4 sin π₯/2 cos π₯/2 = 1/4 Γ 1/2 sin π₯/2 cos π₯/2 = 1/8 We are given sin x , lets first calculate cos x We know that sin2 x + cos2 x = 1 (1/4)2 + cos2 x = 1 1/16 + cos2 x =1 cos2 x = 1 β 1/16 cos2 x = (16 β 1)/16 cos2 x = 15/16 cos x = Β±β(15/16) = Β± β15/4 cos x = Β± β15/4 Given that x is in llnd Quadrant , so cos x is negative So cos x = ββ15/4 Now, cos x = 2 cos2 π₯/2 β 1 cos x = 2 cos2 π₯/2 β 1 ββ15/4 = 2cos2 π₯/2 β 1 1 ββ15/4 = 2cos2 π₯/2 (4 β β15)/4 = 2cos2 π₯/2 2cos2 π₯/2 = (4 β β15)/4 cos2 π₯/2 = (4 β β15)/4 Γ 1/2 cos2 π₯/2 =(4 β β15)/(4 Γ 2) cos π₯/2 = Β± β((4 β β15)/(4 Γ 2)) = Β± β((4 β β15)/(4 Γ 2)Γ 2/2) = Β± β((2(4 β β15))/(4 Γ4)) = Β± β((8 β2β15))/4 Since π₯/2 lie on the lst Quadrant , cos π₯/2 is positive in the lst Quadrant So cos π₯/2 = β((8 β2β15))/4 Also, from (1) sin π₯/2 cos π₯/2 = 1/8 Putting value of cos π₯/2 (sin π₯/2) Γ β((8 β2β15))/4 = 1/8 (sin π₯/2) = 1/8 Γ 4/β((8 β2β15)) (sin π₯/2) = 1/2 Γ 1/β((8 β2β15)) (sin π₯/2) = 1/2 Γ 1/β((8 β2β15) Γ ((8 + 2β15))/((8 + 2β15) )) (sin π₯/2) = 1/2 Γ 1/β( ((8 β2β15) Γ (8 + 2β15))/((8 + 2β15) )) = 1/2 Γ β(((8 + 2β15))/((8 β2β15) Γ (8 + 2β15) )) = 1/2 Γ β(((8 + 2β15))/((82 β (2β15)2 )) = 1/2 Γ β(((8 + 2β15))/((64 β (2)2 (β15)2 )) = 1/2 Γ β(((8 + 2β15))/((64 β 4 Γ 15) )) = 1/2 Γ β(((8 + 2β15))/((64 β60) )) = 1/2 Γ β(((8 + 2β15))/((4) )) = 1/2 Γ β((8 + 2β15) )/β4 = 1/2 Γ β((8 + 2β15) )/2 = β((8 + 2β15) )/4 β΄ sin π₯/2 = β((8 + 2β15) )/4 We know that tan x = sinβ‘π₯/πππ β‘π₯ Replacing x with π₯/2 tan π₯/2 = π ππβ‘γ π₯/2γ/γπππ γβ‘γπ₯/2γ = (β((8 + 2β15) )/4)/(β((8 β2β15))/4) = β((8 + 2β15) )/β((8 β 2β15) ) = β(((8 + 2β15))/((8 β 2β15) )Γ((8 + 2β15))/((8 + 2β15) )) = β((8 + 2β15)2/(8 β 2β15)(8 + 2β15) ) = β((8 + 2β15)2/((82 β(2β15)2) )) = β((8 + 2β15)2/((64 β (4 Γ 15)) )) = β((8 + 2β15)2/((64 β 60) )) = β((8 + 2β15)2/4) = β((8 + 2β15)2/22) = β((((8 + 2β15))/2)^2 ) = ((8 + 2β15))/2 = 2(4 + β15)/2 = (4 + β15) Hence, tan π₯/2 = (4 + β15)

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CA Maninder Singh

CA Maninder Singh is a Chartered Accountant for the past 8 years. He provides courses for Practical Accounts, Taxation and Efiling at teachoo.com .