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Example 28 - If tan x = 3/4 , find sin x/2, cos x/2, tan x/2

Example 28 - Chapter 3 Class 11 Trigonometric Functions - Part 2
Example 28 - Chapter 3 Class 11 Trigonometric Functions - Part 3 Example 28 - Chapter 3 Class 11 Trigonometric Functions - Part 4 Example 28 - Chapter 3 Class 11 Trigonometric Functions - Part 5 Example 28 - Chapter 3 Class 11 Trigonometric Functions - Part 6 Example 28 - Chapter 3 Class 11 Trigonometric Functions - Part 7 Example 28 - Chapter 3 Class 11 Trigonometric Functions - Part 8 Example 28 - Chapter 3 Class 11 Trigonometric Functions - Part 9

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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

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.