Example 16 - Chapter 3 Class 11 Trigonometric Functions (Term 2)
Last updated at Feb. 13, 2020 by
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
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Chapter 3 Class 11 Trigonometric Functions (Term 2)
Serial order wise
- Ex 3.1
- Ex 3.2
- Ex 3.3
- Ex 3.4
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About the Author
Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.
