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

  1. Chapter 3 Class 11 Trigonometric Functions
  2. Serial order wise
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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 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 – 1/16 cos2 x = (16 βˆ’ 1)/16 cos2 x = 15/16 cos x = ±√(15/16) = Β± √15/4 cos x = Β± √15/4 Given that x is in llnd Quadrant , so cos x is negative So cos x = βˆ’βˆš15/4 Now, cos x = 2 cos2 π‘₯/2 – 1 cos x = 2 cos2 π‘₯/2 – 1 βˆ’βˆš15/4 = 2cos2 π‘₯/2 – 1 1 βˆ’βˆš15/4 = 2cos2 π‘₯/2 (4 βˆ’ √15)/4 = 2cos2 π‘₯/2 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’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)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(4 + √15)/2 = (4 + √15) Hence, tan π‘₯/2 = (4 + √15)

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