Slide63.JPG

Slide64.JPG
Slide65.JPG Slide66.JPG Slide67.JPG

Subscribe to our Youtube Channel - https://you.tube/teachoo

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

Transcript

Example 27 find the value of tan πœ‹/8. tan πœ‹/8 Putting Ο€ = 180Β° = tan (180Β°)/8 = tan (45Β°)/2 We find tan (45Β°)/2 using tan 2x formula tan 2x = (2 tan⁑π‘₯)/(1 βˆ’π‘‘π‘Žπ‘›2π‘₯) Putting x = (45Β°)/2 tan ("2 Γ— " (45Β°)/2) = (2 tan⁑〖 (45Β°)/2γ€—)/(1 βˆ’π‘‘π‘Žπ‘›2 (45Β°)/2) tan 45Β° = (2 tan⁑〖 (45Β°)/2γ€—)/(1 βˆ’π‘‘π‘Žπ‘›2 (45Β°)/2) tan 45Β° = (2 tan⁑〖 (45Β°)/2γ€—)/(1 βˆ’π‘‘π‘Žπ‘›2 (45Β°)/2) 1 = (2 tan⁑〖 (45Β°)/2γ€—)/(1 βˆ’π‘‘π‘Žπ‘›2 (45Β°)/2) 1 – tan2 (45Β°)/2 = 2tan (45Β°)/2 (As tan 45Β° = 1) Let tan (πŸ’πŸ“Β°)/𝟐 = x So, our equation becomes 1 – x2 = 2x 0 = 2x + x2 – 1 x2 + 2x – 1 = 0 The above equation is of the form ax2 + bx + c = 0 where a = 1, b = 2, c = βˆ’1 Solution are x = (βˆ’ 𝑏 Β± √(𝑏2 βˆ’4π‘Žπ‘) )/2π‘Ž = (βˆ’ 2 Β± √((βˆ’2)2 βˆ’ 4 Γ— 1 Γ— (βˆ’1)) )/(2 Γ— 1) = (βˆ’2 Β± √(4 + 4))/2 = (βˆ’2 Β± √8)/2 = (βˆ’2 Β± √(2 Γ— 2 Γ— 2))/2 = (βˆ’2 Β± 2√2)/2 = (2 ( βˆ’1 ±√2 ))/2 = –1 Β± √2 Thus, x = –1 Β± √2 tan (45Β°)/2 = –1 Β± √2 But tan (45Β°)/2 = –1 – √2 is not possible as (45Β°)/2 = 22.5Β° lies in first quadrant & tan is positive in first quadrant Therefore, tan (45Β°)/2 = βˆ’1 + √2 i.e. tan 𝝅/πŸ– = √𝟐 – 1

About the Author

Davneet Singh's photo - Teacher, Computer Engineer, Marketer
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.