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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⁑〖7π‘₯ + sin⁑〖5π‘₯) + (sin⁑〖9π‘₯ + sin⁑〖3π‘₯)γ€— γ€— γ€— γ€—)/((cos⁑〖7π‘₯ + π‘π‘œπ‘  5π‘₯) + (cos⁑〖9π‘₯ + cos⁑〖3π‘₯)γ€— γ€— γ€— ) = tan 6x Taking L.H.S ((sin⁑〖7π‘₯ + sin⁑〖5π‘₯) + (sin⁑〖9π‘₯ + sin⁑〖3π‘₯)γ€— γ€— γ€— γ€—)/((cos⁑〖7π‘₯ + π‘π‘œπ‘  5π‘₯) + (cos⁑〖9π‘₯ + cos⁑〖3π‘₯)γ€— γ€— γ€— ) Lets solve numerator and Denominator separately Solving numerator (sin 7x + sin 5x) + ( sin 9x + sin 3x) = 2 sin ((7π‘₯ + 5π‘₯)/2) cos ((7π‘₯ βˆ’ 5π‘₯)/2) + 2sin ((9π‘₯ +3π‘₯)/2) cos ((9π‘₯ βˆ’3π‘₯)/2) = 2 sin (12π‘₯/2) . cos (2π‘₯/2) + 2sin (12π‘₯/2) cos (6π‘₯/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 ((7π‘₯ + 5π‘₯)/2) cos ((7π‘₯ βˆ’ 5π‘₯)/2) + 2 cos ((9π‘₯ +3π‘₯)/2) cos ((9π‘₯ βˆ’3π‘₯)/2) = 2 cos (12π‘₯/2) . cos (2π‘₯/2) + 2 cos (12π‘₯/2) cos (6π‘₯/2) = 2 cos 6x . cos x + 2 cos 6x . cos 3x = 2 cos 6x (cos x + cos 3x) Now solving L.H.S ((sin⁑〖7π‘₯ + sin⁑〖5π‘₯) + (sin⁑〖9π‘₯ + sin⁑〖3π‘₯)γ€— γ€— γ€— γ€—)/((cos⁑〖7π‘₯ + π‘π‘œπ‘  5π‘₯) + (cos⁑〖9π‘₯ + cos⁑〖3π‘₯)γ€— γ€— γ€— ) = (2 sin⁑6π‘₯ (cos⁑π‘₯ + cos⁑3π‘₯))/(2 cos⁑6π‘₯ (cos⁑π‘₯ + cos⁑3π‘₯)) = (sin⁑6π‘₯ )/(cos⁑6π‘₯ ) = tan 6x = R.H.S Hence L.H.S = R.H.S Hence proved

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