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Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class


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° < 𝒙/𝟐 < 90Β° So, 𝒙/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 𝒙/𝟐 cos 𝒙/𝟐 = 𝟏/πŸ’ sin π‘₯/2 cos π‘₯/2 = 1/4 Γ— 1/2 sin 𝒙/𝟐 cos 𝒙/𝟐 = 𝟏/πŸ– 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± βˆšπŸπŸ“/πŸ’ Given that x is in llnd Quadrant So cos x is negative ∴ cos x = (βˆ’βˆšπŸπŸ“)/πŸ’ We know that cos 2x = 2cos2 x – 1 Replacing x by π‘₯/2 cos x = 2 cos2 π‘₯/2 – 1 2cos2 π‘₯/2 – 1 = βˆ’βˆš15/4 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) cos2 π‘₯/2 = (4 βˆ’ √15)/8 cos π‘₯/2 = Β± √((4 βˆ’ √15)/8) = Β± √((4 βˆ’ √15)/8Γ— 2/2) = Β± √((2(4 βˆ’ √15))/16) = Β± √((8 βˆ’2√15))/4 Since π‘₯/2 lie on the lst Quadrant, cos 𝒙/𝟐 is positive in the lst Quadrant So, cos π‘₯/2 = √((8 βˆ’2√15))/4 So, cos 𝒙/𝟐 = √((πŸ– βˆ’πŸβˆšπŸπŸ“))/πŸ’ 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) )) "sin " π‘₯/2 Γ— √((8 βˆ’2√15))/4 = 1/8 "sin " π‘₯/2 = 1/8 Γ— 4/√((8 βˆ’2√15)) "sin " 𝒙/𝟐 = 1/2 Γ— 1/√((8 βˆ’2√15)) = 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 𝒙/𝟐 = √((πŸ– + πŸβˆšπŸπŸ“) )/πŸ’ We know that tan x = sin⁑π‘₯/π‘π‘œπ‘ β‘π‘₯ Replacing x with π‘₯/2 tan 𝒙/𝟐 = π’”π’Šπ’β‘γ€– 𝒙/πŸγ€—/〖𝒄𝒐𝒔 〗⁑〖𝒙/πŸγ€— = (√((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 ) = ((πŸ– + πŸβˆšπŸπŸ“))/𝟐 = 2(4 + √15)/2 \ Hence, tan 𝒙/𝟐 = (πŸ’ + βˆšπŸπŸ“)

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.