1. Chapter 3 Class 11 Trigonometric Functions
2. Serial order wise

Transcript

Example 26 Prove that cos 2x cos ๐ฅ/2 โ cos 3x cos 9๐ฅ/2 = sin 5x sin 5๐ฅ/2 Taking L.H.S Now, cos 2x cos ๐ฅ/2 โ cos 3x cos 9๐ฅ/2 Putting values = 1/2 ("cos " (5x/2)" + cos " (3x/2)) โ 1/2 ("cos " (15๐ฅ/2)" + cos " ((โ3๐ฅ)/2)) = 1/2 ("cos " (5x/2)" + cos " (3x/2)) โ 1/2 ("cos " (15๐ฅ/2)" + cos " (3๐ฅ/2)) = 1/2 ("cos " (5x/2)" + cos " (3x/2)"โ cos " (15๐ฅ/2)"โ cos " (3๐ฅ/2)) = 1/2 ("cos " (5x/2)"โ cos " (15๐ฅ/2)"+ cos " (3x/2)"โ cos " (3๐ฅ/2)) = 1/2 ("cos " (5x/2)"โ cos " (15๐ฅ/2)"+ " 0) = 1/2 ("cos " (5x/2)"โ cos " (15๐ฅ/2)) = 1/2 ("โ 2 sin " ((5๐ฅ/2 +15/2 ๐ฅ)/2)" . sin " ((5๐ฅ/2 โ 15/2 ๐ฅ)/2)) = 1/2 ("โ 2 sin " ((5๐ฅ + 15๐ฅ)/(2 ร 2))" . sin " ((5๐ฅ โ 15๐ฅ)/(2 ร 2))) = 1/2 ("โ 2 sin " (20๐ฅ/4)" . sin " (( โ10๐ฅ)/4)) = 1/2 ("โ 2 sin " (5๐ฅ)" . sin " (( โ5๐ฅ)/2)) = 1/2 ("โ 2 sin " (5๐ฅ)" ร " (โ "sin " (( 5๐ฅ)/2))) = 1/2 ("2 sin " (5๐ฅ)" sin " (( 5๐ฅ)/2)) = sin (5๐ฅ) sin (( 5๐ฅ)/2) = R.H.S Hence L.H.S. = R.H.S. Hence proved