Misc 6 - Prove sin 7x + sin 5x + sin 9x + sin 3x - Class 11 - cos x + cos y formula

  1. Chapter 3 Class 11 Trigonometric Functions
  2. Serial order wise
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Misc 6 Prove that ((sin 7 + sin 5 ) + (sin 9 + sin 3 ) )/((cos 7 + 5 ) + (cos 9 + cos 3 ) ) = tan 6x Taking L.H.S ((sin 7 + sin 5 ) + (sin 9 + sin 3 ) )/((cos 7 + 5 ) + (cos 9 + cos 3 ) ) Lets solve numerator and Denominator separately Solving numerator (sin 7x + sin 5x) + ( sin 9x + sin 3x) = 2 sin ((7 + 5 )/2) cos ((7 5 )/2) + 2sin ((9 +3 )/2) cos ((9 3 )/2) = 2 sin (12 /2) . cos (2 /2) + 2sin (12 /2) cos (6 /2) = 2 sin 6x . cos x + 2 sin 6x . cos 3x = 2sin 6x (cos x + cos 3x) Now solving Denominator ( cos 7x + cos 5x) + ( cos 9x + cos 3x) = 2 cos ((7 + 5 )/2) cos ((7 5 )/2) + 2 cos ((9 +3 )/2) cos ((9 3 )/2) = 2 cos (12 /2) . cos (2 /2) + 2 cos (12 /2) cos (6 /2) = 2 cos 6x . cos x + 2 cos 6x . cos 3x = 2 cos 6x (cos x + cos 3x) Now solving L.H.S ((sin 7 + sin 5 ) + (sin 9 + sin 3 ) )/((cos 7 + 5 ) + (cos 9 + cos 3 ) ) = (2 sin 6 (cos + cos 3 ))/(2 cos 6 (cos + cos 3 )) = (sin 6 )/(cos 6 ) = tan 6x = R.H.S Hence L.H.S = R.H.S Hence proved

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.