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  1. Chapter 3 Class 11 Trigonometric Functions
  2. Serial order wise

Transcript

Misc 10 Find sin π‘₯/2, cos π‘₯/2 and tan π‘₯/2 for sin⁑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 sin π‘₯/2 , cos π‘₯/2 and tan π‘₯/2 are positive Given sin x = 1/4 2 sin π‘₯/2 cos π‘₯/2 = 1/4 We know that sin 2x = 2sin x cos x Replacing x with π‘₯/2 sin 2(π‘₯/2) = 2sin π‘₯/2 cos π‘₯/2 sin x = 2 sin π‘₯/2 cos π‘₯/2 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 – 1/16 cos2 x = (16 βˆ’ 1)/16 cos2 x = 15/16 cos x = ±√(15/16) cos x = Β± √15/4 Given that x is in llnd Quadrant So cos x is negative ∴ cos x = (βˆ’βˆš15)/4 Now, cos x = 2 cos2 π‘₯/2 – 1 cos x = 2 cos2 π‘₯/2 – 1 2cos2 π‘₯/2 – 1 = βˆ’βˆš15/4 We know that cos 2x = 2cos2 x – 1 Replacing x by π‘₯/2 cos 2π‘₯/2 = 2 cos2 π‘₯/2 – 1 cos x = 2 cos2 π‘₯/2 – 1 2cos2 π‘₯/2 = (4 βˆ’ √15)/4 cos2 π‘₯/2 = (4 βˆ’ √15)/4 Γ— 1/2 cos2 π‘₯/2 = (4 βˆ’ √15)/(4 Γ— 2) cos π‘₯/2 = Β± √((4 βˆ’ √15)/(4 Γ— 2)) = Β± √((4 βˆ’ √15)/(4 Γ— 2)Γ— 2/2) = Β± √((2(4 βˆ’ √15))/(4 Γ—4)) = Β± √((8 βˆ’2√15))/4 Since π‘₯/2 lie on the lst Quadrant, cos π‘₯/2 is positive in the lst Quadrant So, cos π‘₯/2 = √((8 βˆ’2√15))/4 Also, from (1) sin π‘₯/2 cos π‘₯/2 = 1/8 Putting value of cos π‘₯/2 "sin " π‘₯/2 Γ— √((8 βˆ’2√15))/4 = 1/8 "sin " π‘₯/2 = 1/8 Γ— 4/√((8 βˆ’2√15)) "sin " π‘₯/2 = 1/2 Γ— 1/√((8 βˆ’2√15)) "sin " π‘₯/2 = 1/2 Γ— 1/√((8 βˆ’2√15) Γ— ((8 + 2√15))/((8 + 2√15) )) "sin " π‘₯/2 = 1/2 Γ— 1/√( ((8 βˆ’2√15) Γ— (8 + 2√15))/((8 + 2√15) )) = 1/2 Γ— √(((8 + 2√15))/((8 βˆ’2√15) Γ— (8 + 2√15) )) = 1/2 Γ— √(((8 + 2√15))/((82 βˆ’ (2√15)2 )) = 1/2 Γ— √(((8 + 2√15))/((64 βˆ’ (2)2 (√15)2 )) = 1/2 Γ— √(((8 + 2√15))/((64 βˆ’ 4 Γ— 15) )) = 1/2 Γ— √(((8 + 2√15))/((64 βˆ’ 4 Γ— 15) )) = 1/2 Γ— √(((8 + 2√15))/((64 βˆ’ 60) )) = 1/2 Γ— √(((8 + 2√15))/4) = 1/2 Γ— √((8 + 2√15) )/√4 = 1/2 Γ— √((8 + 2√15) )/2 = √((8 + 2√15) )/4 ∴ sin π‘₯/2 = √((8 + 2√15) )/4 We know that tan x = sin⁑π‘₯/π‘π‘œπ‘ β‘π‘₯ Replacing x with π‘₯/2 tan π‘₯/2 = 𝑠𝑖𝑛⁑〖 π‘₯/2γ€—/γ€–π‘π‘œπ‘  〗⁑〖π‘₯/2γ€— = (√((8 + 2√15) )/4)/(√((8 βˆ’2√15))/4) = √((8 + 2√15) )/√((8 βˆ’ 2√15) ) = √(((8 + 2√15))/((8 βˆ’ 2√15) )Γ—((8 + 2√15))/((8 + 2√15) )) = √((8 + 2√15)2/(8 βˆ’ 2√15)(8 + 2√15) ) = √(((8 + 2√15))/((8 βˆ’ 2√15) )Γ—((8 + 2√15))/((8 + 2√15) )) = √((8 + 2√15)2/(8 βˆ’ 2√15)(8 + 2√15) ) = √((8 + 2√15)2/((82 βˆ’(2√15)2) )) = √((8 + 2√15)2/((64 βˆ’ (4 Γ— 15)) )) = √((8 + 2√15)2/((64 βˆ’ 60) )) = √((8 + 2√15)2/4) = √((8 + 2√15)2/22) = √((((8 + 2√15))/2)^2 ) = √((((8 + 2√15))/2)^2 ) = ((8 + 2√15))/2 = 2(4 + √15)/2 = (4 + √15) Hence, tan 𝒙/𝟐 = (πŸ’ + βˆšπŸπŸ“)

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.