Misc 7 - Chapter 3 Class 11 Trigonometric Functions (Term 2)

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Misc 7 Prove that: sin 3x + sin2x – sin x = 4 sin x cos 𝑥/2 cos 3𝑥/2 Solving L.H.S sin 3x + sin 2x − sin x = sin 3x + (sin 2x – sin x ) = sin 3x + 2cos ((2𝑥 + 𝑥)/2) . sin ((2𝑥−𝑥)/2) = sin 3x + 2cos (3𝑥/2) sin 𝑥/2 Using sin x – sin y = 2 cos (𝑥 + 𝑦)/2 sin (𝑥 − 𝑦)/2 Putting x = 2x & y = x , Rough As (3𝑥 + 𝑥)/2 = 4𝑥/2 = 2x & (2𝑥 + 𝑥)/2 = 3𝑥/2 As 3𝑥/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 𝟑𝐱/𝟐 cos 𝟑𝐱/𝟐 = 2 sin 3𝑥/2 cos 3𝑥/2 + ["2 cos " 3𝑥/2 " sin " 𝑥/2] = 2 cos 3𝑥/2 ["sin " 3𝑥/2 " + sin " 𝑥/2] Using sin x + sin y = 2 sin (𝑥 + 𝑦)/2 cos (𝑥 − 𝑦)/2 Putting x = 3𝑥/2 & y = 𝑥/2 , = 2 cos 3𝑥/2 ["2 sin " ((3𝑥/2 " + " 𝑥/2))/2 " . cos " ((3𝑥/2 " − " 𝑥/2))/2] = 2 cos 3𝑥/2 ["2 sin " (((3𝑥 + 𝑥)/2))/2 " . cos " (((3𝑥 − 𝑥)/2))/2] = 2 cos 3𝑥/2 ["2 sin " ((4𝑥/2))/2 " . cos " ((2𝑥/2))/2] = 2 cos 3𝑥/2 ["2 sin " ((2𝑥/1))/2 " . cos " ((𝑥/1))/2] = 2 cos 3𝑥/2 ["2 sin " 2𝑥/2 " . cos " 𝑥/2] = 2 cos 3𝑥/2 ["2 sin " 𝑥" . cos " 𝑥/2] = 4 cos 3𝑥/2 sin 𝑥 cos 𝑥/2 = R.H.S Hence L.H.S = R.H.S Hence proved