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

Transcript

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 by 2 all sides (90Β°)/2 < π‘₯/2 < (180Β°)/2 45Β° < π‘₯/2 < 90Β° So, π‘₯/2 lies in Ist quadrant In 1st quadrant, sin , cos & tan are positive ∴ sin π‘₯/2 , cos π‘₯/2 and tan π‘₯/2 are positive Given tan x = (βˆ’4)/3 We know that tan 2x = (2 π‘‘π‘Žπ‘›β‘π‘₯)/(1 βˆ’ π‘‘π‘Žπ‘›2π‘₯) Replacing x with π‘₯/2 tan (2π‘₯/2) = (2 π‘‘π‘Žπ‘›β‘(π‘₯/2))/(1 βˆ’ π‘‘π‘Žπ‘›2(π‘₯/2) ) tan x = (2 π‘‘π‘Žπ‘›β‘(π‘₯/2))/(1 βˆ’ π‘‘π‘Žπ‘›2(π‘₯/2) ) (2 tan⁑(π‘₯/2))/(1 βˆ’ π‘‘π‘Žπ‘›2(π‘₯/2) ) = βˆ’4/3 βˆ’4/3 = (2 tan⁑(π‘₯/2))/(1 βˆ’ π‘‘π‘Žπ‘›2(π‘₯/2) ) –4(2π‘₯/2) = 3Γ— 2 tan (π‘₯/2) –4 Γ— 1 – (–4) Γ— tan2 (π‘₯/2) = 6 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 𝒙/𝟐 by a Our equation becomes –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 = βˆ’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 𝒙/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 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 𝒙/𝟐 = βˆšπŸ“ Therefore, cos 𝒙/𝟐 = 𝟏/βˆšπŸ“ Now, 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) sin π‘₯/2 = Β± √4/√5 sin π‘₯/2 = Β± 2/√5 sin π‘₯/2 = Β± 2/√5 Γ— √5/√5 sin π‘₯/2 = Β± (2√5)/5 Since π‘₯/2 lies on the 1st Quadrant sin π‘₯/2 is positive in the 1st Quadrant So, sin 𝒙/𝟐 = (πŸβˆšπŸ“)/πŸ“ Therefore, tan π‘₯/2 = 2 , cos 𝒙/𝟐 = βˆšπŸ“/πŸ“ & sin 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.