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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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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.