Last updated at Dec. 24, 2019 by Teachoo

Transcript

E ample 29 Prove that cos2 +cos2 ( +" " /3) + cos2 ( " " /3) = 3/2 Lets first calculate all 3 terms separately We know that cos 2x = 2cos2 x 1 cos 2x + 1 = 2cos2 x 2 + 1 /2 = cos2 x So, cos2x = cos 2x + 1 /2 Replacing x with x + /3 is about cos2 (x + /3) = cos 2( + /3) + 1 /2 = cos (2 + 2 /3) + 1 /2 Similarly , replacing x with x + /3 in cos2x = 2 + 1 /2 cos2 (x /3) = cos 2( /3) + 1 /2 = cos ( 2 2 /3) + 1 /2 Now we solve L.H.S cos2x + cos2 ( + 2 /3) + cos2 ( 2 /3) = (1 + cos 2 )/2 + (1 + cos ( 2 + 2 /3))/2 + 1 + cos ( 2 2 /3)/2 = 1/2 [1+cos 2 + 1+ cos ( 2 +2 /3) + 1+ cos ( 2 2 /3) ] = 1/2 [ 1 + 1 + 1 + cos 2x + cos ( 2 +2 /3) + cos ( 2 2 /3)] = 1/2 [ 3 + cos 2x + cos ( 2 + 2 /3) + cos ( 2 2 /3)] = 1/2 ["3 + cos 2x + 2cos " ( (2 + 2 /3 + 2 2 /3)/2)" ." cos ( (2 + 2 /3 (2 +2 /3))/2) ] = 1/2 ["3 + cos 2x + 2cos " ( (2 + 2 + 2 /3 2 /3)/2)" ." cos ( (2 2 + 2 /3 + 2 /3)/2) ] = 1/2 ["3 + cos 2x + 2cos " ( (4 +0)/2)" ." cos ( (0 + 4 /3)/2) ] = 1/2 [ 3 + cos 2x + 2cos ( (4 +0)/2) . cos ( (0 + 4 /3)/2) = 1/2 ["3 + cos 2x + 2cos " ( 4 /2)" ." cos ( (4 /3)/2) ] = 1/2 ["3 + cos 2x + 2cos 2x cos " 2 /3] = 1/2 ["3 + cos 2x + 2cos 2x cos " ( /3)] = 1/2 ["3 + cos 2x + 2cos 2x " ( "cos " ( /3))] = 1/2 [ 3 + cos 2x + 2cos 2x ( 1/2)] = 1/2 [ 3 + cos 2x 2 1/2 cos 2x ] = 1/2 [ 3 + cos 2x cos 2x ] = 1/2 [ 3 + 0] = 3/2 = R.H.S. Hence, L.H.S. = R.H.S. Hence proved

Chapter 3 Class 11 Trigonometric Functions

Concept wise

- Radian measure - Conversion
- Arc length
- Finding Value of trignometric functions, given other functions
- Finding Value of trignometric functions, given angle
- (x + y) formula
- 2x 3x formula - Proving
- 2x 3x formula - Finding value
- cos x + cos y formula
- 2 sin x sin y formula
- Finding Principal solutions
- Finding General Solutions
- Sine and Cosine Formula

About the Author

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.