# sin1-sin2-2-sin3-3-sin4-4-1-2-

Question Number 134439 by Dwaipayan Shikari last updated on 03/Mar/21
$${sin}\mathrm{1}β\frac{{sin}\mathrm{2}}{\mathrm{2}}+\frac{{sin}\mathrm{3}}{\mathrm{3}}β\frac{{sin}\mathrm{4}}{\mathrm{4}}+…=\frac{\mathrm{1}}{\mathrm{2}} \\$$
Answered by mnjuly1970 last updated on 03/Mar/21
$$\:\:\:\boldsymbol{\Omega}=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(β\mathrm{1}\right)^{{n}β\mathrm{1}} {sin}\left({nx}\right)}{{n}}=\:\:\: \\$$$$={im}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(β\mathrm{1}\right)^{{n}β\mathrm{1}} \left({e}^{{ix}} \right)^{{n}} }{{n}}={im}\left({ln}\left(\mathrm{1}+{e}^{{ix}} \right)\right) \\$$$$\:\:\:\underset{{z}={x}+{iy}} {\overset{{ln}\left({z}\right)} {=}}\:{ln}\left(\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }\right)+{itan}^{β\mathrm{1}} \left(\frac{{y}}{{x}}\right) \\$$$$\:\:\:\:={im}\left({ln}\sqrt{\left(\mathrm{1}+{cos}\left({x}\right)\right)^{\mathrm{2}} +{sin}^{\mathrm{2}} \left({x}\right)\:}\:+{itan}^{β\mathrm{1}} \left(\frac{{sin}\left({x}\right)}{\mathrm{1}+{cos}\left({x}\right)}={tan}\left(\frac{{x}}{\mathrm{2}}\right)\right)\right. \\$$$$=\frac{{x}}{\mathrm{2}} \\$$$${x}:=\mathrm{1} \\$$$$\:\:\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(β\mathrm{1}\right)^{{n}β\mathrm{1}} \frac{{sin}\left({n}\right)}{{n}}=\frac{\mathrm{1}}{\mathrm{2}}\:…\checkmark\checkmark \\$$
Commented by Dwaipayan Shikari last updated on 03/Mar/21
$${Great}\:{sir}!\:{It}\:{is}\:{simple}\:{and}\:{nice}. \\$$$$\frac{{x}}{\mathrm{2}}={sinx}β\frac{{sin}\mathrm{2}{x}}{\mathrm{2}}+\frac{{sin}\mathrm{3}{x}}{\mathrm{3}}β\frac{{sin}\mathrm{4}{x}}{\mathrm{4}}+…\:\:\left(\mathrm{0}<{x}<\pi\right) \\$$$${But}\:,{Differentiating}\:{respect}\:{to}\:{x} \\$$$$\frac{\mathrm{1}}{\mathrm{2}}={cosx}β{cos}\mathrm{2}{x}+{cos}\mathrm{3}{x}β{cos}\mathrm{4}{x}+….\:\:\:\left({Independent}\:{of}\:{x}!\right) \\$$$${Is}\:{it}\:{true}? \\$$$${If}\:{we}\:{take}\:{x}=\mathrm{0} \\$$$$\frac{\mathrm{1}}{\mathrm{2}}=\mathrm{1}β\mathrm{1}+\mathrm{1}β\mathrm{1}+\mathrm{1}β\mathrm{1}+….\: \\$$$${But}\:\mathrm{0}<{x}<\pi \\$$
Commented by mnjuly1970 last updated on 03/Mar/21
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