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Question-120619




Question Number 120619 by help last updated on 01/Nov/20
Answered by TANMAY PANACEA last updated on 01/Nov/20
16sin^5 θ  16(((e^(iθ) −e^(−iθ) )/(2i)))^5  e^(iθ) =cosθ+isinθ   e^(−iθ) =cosθ−isinθ  (1/(2i))(e^(5iθ) −5c_1 e^(i4θ) .e^(−iθ) +5c_2 e^(i3θ) .e^(−i2θ) −5c_3 e^(i2θ) e^(−i3θ) +5c_4 e^(iθ) e^(−i4θ) −e^(−5iθ) )  =(((e^(5iθ) −e^(−i5θ) )/(2i)))−5(((e^(i3θ) −e^(−i3θ) )/(2i)))+((5×4)/2)(((e^(iθ) −e^(−iθ) )/(2i)))  =sin5θ−5sin3θ+10sinθ
$$\mathrm{16}{sin}^{\mathrm{5}} \theta \\ $$$$\mathrm{16}\left(\frac{{e}^{{i}\theta} −{e}^{−{i}\theta} }{\mathrm{2}{i}}\right)^{\mathrm{5}} \:{e}^{{i}\theta} ={cos}\theta+{isin}\theta\:\:\:{e}^{−{i}\theta} ={cos}\theta−{isin}\theta \\ $$$$\frac{\mathrm{1}}{\mathrm{2}{i}}\left({e}^{\mathrm{5}{i}\theta} −\mathrm{5}{c}_{\mathrm{1}} {e}^{{i}\mathrm{4}\theta} .{e}^{−{i}\theta} +\mathrm{5}{c}_{\mathrm{2}} {e}^{{i}\mathrm{3}\theta} .{e}^{−{i}\mathrm{2}\theta} −\mathrm{5}{c}_{\mathrm{3}} {e}^{{i}\mathrm{2}\theta} {e}^{−{i}\mathrm{3}\theta} +\mathrm{5}{c}_{\mathrm{4}} {e}^{{i}\theta} {e}^{−{i}\mathrm{4}\theta} −{e}^{−\mathrm{5}{i}\theta} \right) \\ $$$$=\left(\frac{{e}^{\mathrm{5}{i}\theta} −{e}^{−{i}\mathrm{5}\theta} }{\mathrm{2}{i}}\right)−\mathrm{5}\left(\frac{{e}^{{i}\mathrm{3}\theta} −{e}^{−{i}\mathrm{3}\theta} }{\mathrm{2}{i}}\right)+\frac{\mathrm{5}×\mathrm{4}}{\mathrm{2}}\left(\frac{{e}^{{i}\theta} −{e}^{−{i}\theta} }{\mathrm{2}{i}}\right) \\ $$$$={sin}\mathrm{5}\theta−\mathrm{5}{sin}\mathrm{3}\theta+\mathrm{10}{sin}\theta \\ $$
Commented by peter frank last updated on 01/Nov/20
good
$$\mathrm{good} \\ $$

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