Example 27 - Chapter 3 Class 11 Trigonometric Functions
Last updated at May 29, 2018 by Teachoo
Transcript
Example, 27 find the value of tan " " /8 . tan /8 Putting = 180 = tan (180 )/8 = tan (45 )/2 We find tan (45 )/2 using tan 2x formula tan 2x = (2 tan )/(1 2 ) Putting x = (45 )/2 tan (2 (45 )/2) = (2 tan (45 )/2 )/(1 2 (45 )/2) tan 45 = (2 tan (45 )/2 )/(1 2 (45 )/2) 1 = (2 tan (45 )/2 )/(1 2 (45 )/2) 1 tan2 (45 )/2 = 2tan (45 )/2 Putting tan (45 )/2 = x 1 x2 = 2x 0 = 2x + x2 1 0 = x2 + 2x 1 x2 + 2x 1 = 0 The above equation is of the form ax2 + bx + c = 0 Here a = 1, b = 2, c = -1 Solution are x = ( ( 2 4 ) )/2 = ( 2 (( 2)2 4 1 ( 1)) )/(2 1) = ( 2 (4 + 4))/2 = ( 2 8)/2 = ( 2 (2 2 2))/2 = ( 2 2 2)/2 = (2 ( 1 2 ))/2 = 1 2 x = 1 2 tan (45 )/2 = 1 2 But tan (45 )/2 = 1 2 is not possible as (45 )/2 = 22.5 lies in first quadrant & tan is positive in first quadrant tan (45 )/2 = - 1 + 2 So, tan /8 = 2 1
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Example 27 Important You are here
Example 28 Important
Example 29
Chapter 3 Class 11 Trigonometric Functions
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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.