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7 September, 10:33
Express cos 5x in terms of sine multiples of x
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Ezekiel Schmitt
7 September, 13:45
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cos x = [ e^ (ix) - e^ (-ix) ] / 2i along with [ A - B ]^5 =
(A^5 - B^5) - 5 AB (A^3 - B^3) + 10 A²B² (A - B) ... A = e^ (ix), B = e^ (-ix)
[ cos x ] ^5 = (1/32i) { [ e^ (5ix) - e^ (-5x) ] - 5 [ (e^ (3ix) - e^ (-3ix) ] + 10 [ e^ (ix) - e^ (-ix) ] }
= (1 / 16) { cos 5x - 5 cos 3x + 10 cos x } ... much faster
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