Example 16 - Chapter 3 Class 11 Trigonometric Functions (Term 2)
Last updated at Feb. 13, 2020 by Teachoo
Examples
Example 1
Example 2 Important
Example 3
Example 4
Example 5 Important
Example 6 Important
Example 7 Important
Example 8
Example 9 Important
Example 10
Example 11 Important
Example 12
Example 13
Example 14
Example 15
Example 16 Important You are here
Example 17 Important
Example 18
Example 19
Example 20 Deleted for CBSE Board 2023 Exams
Example 21 Deleted for CBSE Board 2023 Exams
Example 22 Important Deleted for CBSE Board 2023 Exams
Example 23 Deleted for CBSE Board 2023 Exams
Example 24 Important Deleted for CBSE Board 2023 Exams
Example 25 Important
Example 26
Example 27 Important
Example 28 Important
Example 29 Important
Chapter 3 Class 11 Trigonometric Functions
Serial order wise
- Ex 3.1
- Ex 3.2
- Ex 3.3
- Ex 3.4
-
Examples
- Miscellaneous
Transcript
Example 16 Prove that πππ β‘γ7π₯ + πππ β‘5π₯ γ/π ππβ‘γ7π₯ β π ππβ‘5π₯ γ = cot x Taking L.H.S. We solve cos 7x + cos 5x & sin 7x β sin 5x separately cos x + cos y = 2 cos (π₯ + π¦)/2 cos (π₯ β π¦)/2 Putting x = 7x & y = 5x cos 7x + cos 5x = 2 cos ((7π₯ + 5π₯)/2) cos ((7π₯ β 5π₯)/2) = 2 cos (12π₯/2) cos (2π₯/2) = 2 cos 6x cos x sin x β sin y = 2 cos (π₯ + π¦)/2 sin (π₯ β π¦)/2 Putting x = 7x & y = 5x sin 7x β sin 5x = 2 cos ((7π₯ + 5π₯)/2) sin((7π₯ β 5π₯)/2) = 2 cos (12π₯/2) sin (2π₯/2) = 2 cos 6x sin x Now γπππ γβ‘γ7π₯ + πππ β‘5π₯ γ/π ππβ‘γ7π₯ β π ππβ‘5π₯ γ = (2 γ πππ γβ‘γ6x πππ β‘π₯ γ)/(2 πππ β‘γ 6π₯ siπβ‘π₯ γ ) = πππ β‘γ π₯γ/π ππβ‘γ π₯γ = cot x = R.H.S. Hence L.H.S. = R.H.S. Hence proved
