sin5(x)=(eix − e−ix2i)5=116 ∗ 2i ∗ (eix − e−ix)5=116 ∗ 2i ∗ (e5ix−5e3ix+10eix−10e−ix+5e−3ix−e−5ix)==116(sin 5x − 5sin 3x + 10sin x)==116sin 5x − 516sin 3x + 58sin xsin^5(x) = (\frac {e^{ix}\ -\ e^{-ix}} {2i} )^ 5 = \frac 1 {16\ *\ 2i}\ *\ (e^{ix}\ -\ e^{-ix})^5=\newline \frac 1 {16\ *\ 2i}\ *\ \newline (e ^ {5ix} - 5e^{3ix} + 10e^{ix} - 10e^{-ix} + 5e^{-3ix} - e^{-5ix}) = \newline = \frac 1 {16} (sin\ 5x\ -\ 5sin\ 3x\ +\ 10sin\ x) = \newline = \frac 1 {16}sin\ 5x\ -\ \frac 5 {16}sin\ 3x\ +\ \frac 5 {8}sin\ xsin5(x)=(2ieix − e−ix)5=16 ∗ 2i1 ∗ (eix − e−ix)5=16 ∗ 2i1 ∗ (e5ix−5e3ix+10eix−10e−ix+5e−3ix−e−5ix)==161(sin 5x − 5sin 3x + 10sin x)==161sin 5x − 165sin 3x + 85sin x
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