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Misc 8 - tan x = -4/3, find sin x/2 , cos x/2 and tan x/2 - 2x 3x formula - Finding value

 

 

  1. Chapter 3 Class 11 Trigonometric Functions
  2. Serial order wise
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Misc 8 Find the value of sin π‘₯/2 , cos π‘₯/2 and tan π‘₯/2 in each of the following : tan⁑π‘₯ = – 4/3 , π‘₯ in quadrant II Given that x is in quadrant II So, 90Β° < x < 180Β° Dividing with 2 all sides (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 tan x = βˆ’4/3 tan x = (2 tan⁑(π‘₯/2))/(1 βˆ’ π‘‘π‘Žπ‘›2(π‘₯/2) ) βˆ’4/3 = (2 tan⁑(π‘₯/2))/(1 βˆ’ π‘‘π‘Žπ‘›2(π‘₯/2) ) –4( 1 – tan2 (π‘₯/2)) = 3Γ— 2 tan (π‘₯/2) –4 Γ— 1 – (–4) Γ— tan2 (π‘₯/2) = 6 tan (π‘₯/2) –4 + 4 tan2 (π‘₯/2) = 6 tan (π‘₯/2) –4 + 4 tan2 (π‘₯/2) – 6 tan (π‘₯/2) = 0 Replacing tan π‘₯/2 by a – 4 + 4a2 – 6a = 0 4a2 – 6a – 4 = 0 4a2 – 8a + 2a – 4 = 0 4a(a – 2) + 2 ( a – 2) = 0 (4a + 2) ( a – 2) = 0 Hence 4a + 2 = 0 4a = - 2 a = (βˆ’2)/( 4) a = βˆ’1/2 So, a = βˆ’1/2 or a = 2 Hence, tan π‘₯/2 = βˆ’1/2 or tan π‘₯/2 = 2 Since, π‘₯/2 lies in Ist quadrant tan π‘₯/2 is positive, ∴ tan π‘₯/2 = 2 Now, We know that 1 + tan2 x = sec2 x Replacing x with π‘₯/2 1 + tan2 π‘₯/2 = sec2 π‘₯/2 1 + (2)2 = sec2 π‘₯/2 1 + 4 = sec2 x/2 5 = sec2 π‘₯/2 sec2 π‘₯/2 = 5 sec π‘₯/2 = Β± √5 Since π‘₯/2 lie on the 1st Quadrant, sec π‘₯/2 is positive in the 1st Quadrant So sec π‘₯/2 = √5 sec π‘₯/2 = √5 β‡’ 1/cos⁑〖 π‘₯/2γ€— = √5 β‡’ 1/√5 = cos π‘₯/2 β‡’ cos π‘₯/2 = 1/√5 cos π‘₯/2 = 1/√5 Γ— √5/√5 cos π‘₯/2 = √5/5 We know that sin2x + cos2x = 1 Replacing x with π‘₯/2 sin2 π‘₯/2 + cos2 π‘₯/2 = 1 sin2 π‘₯/2 = 1 – cos2 π‘₯/2 Putting cos π‘₯/2 = √5/5 sin2 π‘₯/2 = 1 – (√5/5)2 sin2 π‘₯/2 = 1 – 5/25 sin2 π‘₯/2 = 1 – 1/5 sin2 π‘₯/2 = (5 βˆ’ 1)/5 sin2 π‘₯/2 = 4/5 sin π‘₯/2 = Β± √(4/5) = Β± √4/√5 = Β± 2/√5 = Β± 2/√5 Γ— √5/√5 = Β± (2√5)/5 Since π‘₯/2 lie on the 1st Quadrant , sin π‘₯/2 is positive in the 1st Quadrant So sin π‘₯/2 = (2√5)/5 Hence, tan π‘₯/2 = 2 , cos π‘₯/2 = √5/5 & sin π‘₯/2 = (2√5)/5

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He provides courses for Mathematics from Class 9 to 12. You can ask questions here.
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