Learn All Concepts of Chapter 2 Class 11 Relations and Function - FREE. Check - Trigonometry Class 11 - All Concepts

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

Transcript

Example 28 If tan⁑π‘₯ = 3/4 , "Ο€" < π‘₯ < 3πœ‹/4 , find the value of sin π‘₯/2 , cos π‘₯/2 and tan π‘₯/2 Given that "Ο€" < x < 3πœ‹/2 i.e.180Β° < x < 3/2 Γ— 180Β° i.e. 180Β° < x < 270Β° Dividing by 2 all sides (180Β°)/2 < π‘₯/2 < (270Β°)/2 90Β° < π‘₯/2 < 135Β° So, π‘₯/2 lies in 2nd quadrant In 2nd quadrant, sin is positive, cos & tan are negative sin π‘₯/2 is positive, cos π‘₯/2 and tan π‘₯/2 are negative Given tan x = 3/4 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) ) tan x = (2 tan⁑(π‘₯/2))/(1 βˆ’ π‘‘π‘Žπ‘›2(π‘₯/2) ) 3/4 = (2 tan⁑(π‘₯/2))/(1 βˆ’ π‘‘π‘Žπ‘›2(π‘₯/2) ) 3(1 – tan2 (π‘₯/2) = 4 Γ— 2 tan (π‘₯/2) 3 Γ— 1 – 3 Γ— tan2 (π‘₯/2) = 8 tan (π‘₯/2) 3 Γ— 1 – 3 Γ— tan2 (π‘₯/2) = 8 tan (π‘₯/2) 3 – 3 tan2 (π‘₯/2) = 8 tan (π‘₯/2) 0 = –3 + 3tan2 (π‘₯/2) + 8 tan π‘₯/2 Replacing tan 𝐱/𝟐 by a Our equation becomes 0 = –3 + 3a2 + 8a 3a2 + 8a – 3 = 0 3a2 + 9a – a – 3 = 0 3a (a + 3) – 1 (a + 3) = 0 (3a – 1) (a + 3) = 0 Hence So, a = 1/3 or a = –3 3a – 1 = 0 3a = 1 a = 1/3 a + 3 = 0 a = βˆ’3 Hence, tan π‘₯/2 = 1/3 or tan π‘₯/2 = –3 Since π‘₯/2 lies in IInd quadrant So, tan π‘₯/2 is negative, ∴ tan 𝒙/𝟐 = –3 Now, We know that 1 + tan2 x = sec2 x Replacing x with π‘₯/2 1 + tan2 π‘₯/2 = sec2 π‘₯/2 1 + (–3)2 = sec2 π‘₯/2 1 + 9 = sec2 x/2 1 + 9 = sec2 x/2 10 = sec2 π‘₯/2 sec2 π‘₯/2 = 10 sec π‘₯/2 = Β± √10 Since π‘₯/2 lie on the llnd Quadrant, cos π‘₯/2 is negative in the llnd Quadrant ∴ sec π‘₯/2 is negative in the llnd Quadrant So, sec 𝒙/𝟐 = βˆ’βˆšπŸπŸŽ Therefore, cos 𝒙/𝟐 = (βˆ’πŸ)/√𝟏𝟎 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 = βˆ’1/√10 sin2 π‘₯/2 = 1 – ((βˆ’1)/√10)2 sin2 π‘₯/2 = 1 – 1/10 sin2 π‘₯/2 = (10 βˆ’ 1)/10 sin2 π‘₯/2 = 9/10 sin π‘₯/2 = Β± √(9/10) sin π‘₯/2= Β± √9/√10 sin π‘₯/2 = Β± 3/√10 Since π‘₯/2 lies on the 2nd Quadrant sin π‘₯/2 is positive in the 2nd Quadrant So, sin 𝒙/𝟐 = πŸ‘/√𝟏𝟎 Hence, tan π‘₯/2 = βˆ’3 , cos π‘₯/2 = (βˆ’1)/√10 & sin π‘₯/2 = 3/√10

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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.