Example 19 - Prove cos 2x cos x/2 - cos 3x cos 9x/2 = sin 5x - Examples

part 2 - Example 19 - Examples - Serial order wise - Chapter 3 Class 11 Trigonometric Functions
part 3 - Example 19 - Examples - Serial order wise - Chapter 3 Class 11 Trigonometric Functions part 4 - Example 19 - Examples - Serial order wise - Chapter 3 Class 11 Trigonometric Functions part 5 - Example 19 - Examples - Serial order wise - Chapter 3 Class 11 Trigonometric Functions

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Example 19 Prove that cos 2x cos š‘„/2 – cos 3x cos 9š‘„/2 = sin 5x sin 5š‘„/2 Solving L.H.S Solving cos 2x cos x/2 and cos 3x cos 9š‘„/2 separately cos 2x cos š’™/šŸ Replacing x with 2x and y with š‘„/2 = 1/2 ("cos " ("2x + " x/2)" + cos" ("2x" āˆ’x/2)) = 1/2 ("cos " ((4x + x )/2)" + cos " ((4x āˆ’ x)/2)) = šŸ/šŸ ("cos " (šŸ“š’™/šŸ)" + cos " (šŸ‘š’™/šŸ)) cos 3x cos šŸ—š’™/šŸ Replacing x with 3x and y with 9š‘„/2 = 1/2 ("cos" ("3x + " 9š‘„/2)" + cos " ("3x – " 9š‘„/2)) = 1/2 ("cos " ((6š‘„ + 9š‘„ )/2)" + cos " ((6š‘„ āˆ’ 9š‘„)/2)) = šŸ/šŸ ("cos " (šŸšŸ“š’™/šŸ)" + cos " ((āˆ’šŸ‘š’™)/šŸ)) Now, cos 2x cos š‘„/2 – cos 3x cos 9š‘„/2 Putting values = 1/2 ("cos " (5x/2)" + cos " (3x/2)) – 1/2 ("cos " (15š‘„/2)" + cos " ((āˆ’šŸ‘š’™)/šŸ)) = 1/2 ("cos " (5x/2)" + cos " (3x/2)) – 1/2 ("cos " (15š‘„/2)" + cos " (šŸ‘š’™/šŸ)) = 1/2 ("cos " (5x/2)" + cos " (3x/2)"– cos " (15š‘„/2)"– cos " (3š‘„/2)) = 1/2 ("cos " (5x/2)"– cos " (15š‘„/2)"+ cos " (3x/2)"– cos " (3š‘„/2)) = 1/2 ("cos " (5x/2)"– cos " (15š‘„/2)"+ " 0) = 1/2 ("cos " (šŸ“š’™/šŸ)"– cos " (šŸšŸ“š’™/šŸ)) = 1/2 ("– 2 sin " ((5š‘„/2 +15/2 š‘„)/2)". sin " ((5š‘„/2 āˆ’ 15/2 š‘„)/2)) = āˆ’"sin " ((5š‘„ + 15š‘„)/(2 Ɨ 2))" . sin " ((5š‘„ āˆ’ 15š‘„)/(2 Ɨ 2)) = "– sin " (šŸšŸŽš’™/šŸ’)" . sin " (( āˆ’šŸšŸŽš’™)/šŸ’) = āˆ’ "sin " (5š‘„)" . sin " (( āˆ’5š‘„)/2) = āˆ’ "sin " (5š‘„) Ɨ āˆ’" sin " (( 5š‘„)/2) = sin (šŸ“š’™) sin (( šŸ“š’™)/šŸ) = R.H.S Hence L.H.S. = R.H.S. Hence proved

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