Misc 5 - Prove sin x + sin 3x + sin5x + sin 7x = 4 cos x - Miscellaneous

part 2 - Misc 5 - Miscellaneous - Serial order wise - Chapter 3 Class 11 Trigonometric Functions

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Misc 5 Prove that: sin π‘₯ + sin 3π‘₯ + sin5π‘₯ + sin 7π‘₯ = 4cos π‘₯ cos 2π‘₯ sin 4π‘₯ Solving LHS sin π‘₯ + sin 3π‘₯ + sin5π‘₯ + sin 7π‘₯ = (𝐬𝐒𝐧⁑𝒙+π’”π’Šπ’ πŸ“π’™)+(π¬π’π§β‘πŸ‘π’™+π’”π’Šπ’β‘πŸ•π’™) = 2 sin ((π‘₯ + 5π‘₯)/2) .cos ((π‘₯ βˆ’ 5π‘₯)/2) + 2sin ((3π‘₯ + 7π‘₯)/2) cos ((3π‘₯ βˆ’ 7π‘₯)/2) = 2 sin 3π‘₯ cos (–2π‘₯) + 2sin 5π‘₯ cos (–2π‘₯) = 2 sin 3π‘₯ cos 2π‘₯ + 2sin 5π‘₯ cos 2π‘₯ = 2 cos 2π‘₯ [ sin 3π‘₯ + sin 5π‘₯] = 2cos 2π‘₯ ("2sin " ((3π‘₯ + 5π‘₯)/2)" . cos" ((3π‘₯ βˆ’ 5π‘₯)/2)) = 2cos⁑2π‘₯ [2 sin⁑4π‘₯.cos⁑〖(–π‘₯)γ€—] = πŸ’ πœπ¨π¬β‘πŸπ’™ π¬π’π§β‘πŸ’π’™ πœπ¨π¬β‘π’™ = RH.S. = 2 cos 2π‘₯ [ sin 3π‘₯ + sin 5π‘₯] = 2cos 2π‘₯ ("2sin " ((3π‘₯ + 5π‘₯)/2)" . cos" ((3π‘₯ βˆ’ 5π‘₯)/2)) = 2cos⁑2π‘₯ [2 sin⁑4π‘₯.πœπ¨π¬β‘γ€–(–𝒙)γ€—] = πŸ’ πœπ¨π¬β‘πŸπ’™ π¬π’π§β‘πŸ’π’™ πœπ¨π¬β‘π’™ = RH.S. Hence , L.H.S.= R.H.S. Hence proved

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