Misc 7 - Prove sin 3x + sin2x - sin x = 4 sin x cos x/2 cos 3x/2 - Miscellaneous

part 2 - Misc 7 - Miscellaneous - Serial order wise - Chapter 3 Class 11 Trigonometric Functions
part 3 - Misc 7 - Miscellaneous - Serial order wise - Chapter 3 Class 11 Trigonometric Functions part 4 - Misc 7 - Miscellaneous - Serial order wise - Chapter 3 Class 11 Trigonometric Functions

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Misc 7 Prove that: sin 3x + sin2x – sin x = 4 sin x cos š‘„/2 cos 3š‘„/2 Solving L.H.S sin 3x + sin 2x āˆ’ sin x = sin 3x + (sin 2x – sin x) = sin 3x + 2cos ((2š‘„ + š‘„)/2) . sin ((2š‘„āˆ’š‘„)/2) = sin 3x + 2 cos (šŸ‘š’™/šŸ) sin š’™/šŸ We know that sin 2x = 2 sin x cos x Divide by x by x/2 sin 2x/2 = 2 sin x/2 cos x/2 sin x = 2 sin x/2 cos x/2 Now Replace x by 3x sin 3x = 2 sin šŸ‘š±/šŸ cos šŸ‘š±/šŸ = 2 sin 3š‘„/2 cos 3š‘„/2 + ["2 cos " 3š‘„/2 " sin " š‘„/2] = 2 cos 3š‘„/2 ["sin " šŸ‘š’™/šŸ " + sin " š’™/šŸ] Using sin x + sin y = 2 sin (š‘„ + š‘¦)/2 cos (š‘„ āˆ’ š‘¦)/2 Putting x = 3š‘„/2 & y = š‘„/2 , = 2 cos 3š‘„/2 ["2 sin " ((3š‘„/2 " + " š‘„/2))/2 " . cos " ((3š‘„/2 " āˆ’ " š‘„/2))/2] = 2 cos 3š‘„/2 ["2 sin " (((3š‘„ + š‘„)/2))/2 " . cos " (((3š‘„ āˆ’ š‘„)/2))/2] = 2 cos 3š‘„/2 ["2 sin " ((4š‘„/2))/2 " . cos " ((2š‘„/2))/2] = 2 cos 3š‘„/2 ["2 sin " ((2š‘„/1))/2 " . cos " ((š‘„/1))/2] = 2 cos šŸ‘š’™/šŸ ["2 sin " šŸš’™/šŸ " . cos " š’™/šŸ] = 2 cos 3š‘„/2 ["2 sin " š‘„" . cos " š‘„/2] = 4 cos šŸ‘š’™/šŸ sin š’™ cos š’™/šŸ = R.H.S Hence L.H.S = R.H.S Hence proved

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