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2cos 1/2(2A + 5A).cos 1/2(2A - 5A) + 2cos 1/2(8A + 11A)cos 1/2(8A - 11A) =
2.cos (7A/2) cos (- 3A/2) + 2 cos.(19A/2) cos (- 3A/2) =
2.cos (7A/2) cos ( 3A/2) + 2 cos.(19A/2) cos ( 3A/2) =
2 cos (3A/2) [ cos (7A/2) + cos (19A/2) ] =
2.cos (3A/2) [ 2cos 1/2 (7A/2 + 19A/2) cos 1/2(7A - 19A/2) ]
2.cos (3A/2) [ 2 .cos 1/2(26A/2) cos 1/2(- 12A/2) ] =
2 .cos (3A/2) .2 .cos (13A/2) . cos 3A =
4 . cos (3a/2) .cos (13A/2) .cos 3A ...... terbukti