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Misc 10 - sin x = 1/4, find sin x/2 , cos x/2, tan x/2 - Chapter 3

Misc 10 - Chapter 3 Class 11 Trigonometric Functions - Part 2
Misc 10 - Chapter 3 Class 11 Trigonometric Functions - Part 3 Misc 10 - Chapter 3 Class 11 Trigonometric Functions - Part 4 Misc 10 - Chapter 3 Class 11 Trigonometric Functions - Part 5 Misc 10 - Chapter 3 Class 11 Trigonometric Functions - Part 6 Misc 10 - Chapter 3 Class 11 Trigonometric Functions - Part 7 Misc 10 - Chapter 3 Class 11 Trigonometric Functions - Part 8 Misc 10 - Chapter 3 Class 11 Trigonometric Functions - Part 9 Misc 10 - Chapter 3 Class 11 Trigonometric Functions - Part 10 Misc 10 - Chapter 3 Class 11 Trigonometric Functions - Part 11 Misc 10 - Chapter 3 Class 11 Trigonometric Functions - Part 12 Misc 10 - Chapter 3 Class 11 Trigonometric Functions - Part 13 Misc 10 - Chapter 3 Class 11 Trigonometric Functions - Part 14

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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 has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.