Miscellaneous
Last updated at April 16, 2024 by Teachoo
Misc 5 Prove that: sin ๐ฅ + sin 3๐ฅ + sin5๐ฅ + sin 7๐ฅ = 4cos ๐ฅ cos 2๐ฅ sin 4๐ฅ Solving LHS sin ๐ฅ + sin 3๐ฅ + sin5๐ฅ + sin 7๐ฅ = (๐ฌ๐ข๐งโก๐+๐๐๐ ๐๐)+(๐ฌ๐ข๐งโก๐๐+๐๐๐โก๐๐) = 2 sin ((๐ฅ + 5๐ฅ)/2) .cos ((๐ฅ โ 5๐ฅ)/2) + 2sin ((3๐ฅ + 7๐ฅ)/2) cos ((3๐ฅ โ 7๐ฅ)/2) = 2 sin 3๐ฅ cos (โ2๐ฅ) + 2sin 5๐ฅ cos (โ2๐ฅ) = 2 sin 3๐ฅ cos 2๐ฅ + 2sin 5๐ฅ cos 2๐ฅ = 2 cos 2๐ฅ [ sin 3๐ฅ + sin 5๐ฅ] = 2cos 2๐ฅ ("2sin " ((3๐ฅ + 5๐ฅ)/2)" . cos" ((3๐ฅ โ 5๐ฅ)/2)) = 2cosโก2๐ฅ [2 sinโก4๐ฅ.cosโกใ(โ๐ฅ)ใ] = ๐ ๐๐จ๐ฌโก๐๐ ๐ฌ๐ข๐งโก๐๐ ๐๐จ๐ฌโก๐ = RH.S. = 2 cos 2๐ฅ [ sin 3๐ฅ + sin 5๐ฅ] = 2cos 2๐ฅ ("2sin " ((3๐ฅ + 5๐ฅ)/2)" . cos" ((3๐ฅ โ 5๐ฅ)/2)) = 2cosโก2๐ฅ [2 sinโก4๐ฅ.๐๐จ๐ฌโกใ(โ๐)ใ] = ๐ ๐๐จ๐ฌโก๐๐ ๐ฌ๐ข๐งโก๐๐ ๐๐จ๐ฌโก๐ = RH.S. Hence , L.H.S.= R.H.S. Hence proved