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

Misc 7 - Chapter 3 Class 11 Trigonometric Functions - Part 2
Misc 7 - Chapter 3 Class 11 Trigonometric Functions - Part 3
Misc 7 - Chapter 3 Class 11 Trigonometric Functions - Part 4

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Misc 7 Prove that: sin 3x + sin2x – sin x = 4 sin x cos π‘₯/2 cos 3π‘₯/2 Solving L.H.S sin 3x + sin 2x βˆ’ sin x = sin 3x + (sin 2x – sin x ) = sin 3x + 2cos ((2π‘₯ + π‘₯)/2) . sin ((2π‘₯βˆ’π‘₯)/2) = sin 3x + 2cos (3π‘₯/2) sin π‘₯/2 Using sin x – sin y = 2 cos (π‘₯ + 𝑦)/2 sin (π‘₯ βˆ’ 𝑦)/2 Putting x = 2x & y = x , Rough As (3π‘₯ + π‘₯)/2 = 4π‘₯/2 = 2x & (2π‘₯ + π‘₯)/2 = 3π‘₯/2 As 3π‘₯/2 is in R.H.S. , we take x & 2x 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 πŸ‘π±/𝟐 cos πŸ‘π±/𝟐 = 2 sin 3π‘₯/2 cos 3π‘₯/2 + ["2 cos " 3π‘₯/2 " sin " π‘₯/2] = 2 cos 3π‘₯/2 ["sin " 3π‘₯/2 " + sin " π‘₯/2] Using sin x + sin y = 2 sin (π‘₯ + 𝑦)/2 cos (π‘₯ βˆ’ 𝑦)/2 Putting x = 3π‘₯/2 & y = π‘₯/2 , = 2 cos 3π‘₯/2 ["2 sin " ((3π‘₯/2 " + " π‘₯/2))/2 " . cos " ((3π‘₯/2 " βˆ’ " π‘₯/2))/2] = 2 cos 3π‘₯/2 ["2 sin " (((3π‘₯ + π‘₯)/2))/2 " . cos " (((3π‘₯ βˆ’ π‘₯)/2))/2] = 2 cos 3π‘₯/2 ["2 sin " ((4π‘₯/2))/2 " . cos " ((2π‘₯/2))/2] = 2 cos 3π‘₯/2 ["2 sin " ((2π‘₯/1))/2 " . cos " ((π‘₯/1))/2] = 2 cos 3π‘₯/2 ["2 sin " 2π‘₯/2 " . cos " π‘₯/2] = 2 cos 3π‘₯/2 ["2 sin " π‘₯" . cos " π‘₯/2] = 4 cos 3π‘₯/2 sin π‘₯ cos π‘₯/2 = R.H.S Hence L.H.S = R.H.S Hence proved

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.