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Question Number 129233 by MrJoe last updated on 14/Jan/21

prove that sin 3x=2cos(((3x)/2))sin(((3x)/2))

$${prove}\:{that} \\ $$$${sin}\:\mathrm{3}{x}=\mathrm{2}{cos}\left(\frac{\mathrm{3}{x}}{\mathrm{2}}\right){sin}\left(\frac{\mathrm{3}{x}}{\mathrm{2}}\right) \\ $$

Commented by MrJoe last updated on 15/Jan/21

then if sin(3x)=sin[2(((3x)/2))] form sin2x=2sinx cosx sin3x = 2sin(((3x)/2))cos(((3x)/2))

$${then}\: \\ $$$${if}\:{sin}\left(\mathrm{3}{x}\right)={sin}\left[\mathrm{2}\left(\frac{\mathrm{3}{x}}{\mathrm{2}}\right)\right] \\ $$$${form}\:{sin}\mathrm{2}{x}=\mathrm{2}{sinx}\:{cosx}\: \\ $$$${sin}\mathrm{3}{x}\:=\:\mathrm{2}{sin}\left(\frac{\mathrm{3}{x}}{\mathrm{2}}\right){cos}\left(\frac{\mathrm{3}{x}}{\mathrm{2}}\right) \\ $$

Commented by MJS_new last updated on 15/Jan/21

yes. now prove that sin 2x =2sin x cos x

$$\mathrm{yes}.\:\mathrm{now}\:\mathrm{prove}\:\mathrm{that}\:\mathrm{sin}\:\mathrm{2}{x}\:=\mathrm{2sin}\:{x}\:\mathrm{cos}\:{x} \\ $$

Answered by MJS_new last updated on 14/Jan/21

sin α =((e^(iα) −e^(−iα) )/(2i))∧cos α =((e^(iα) +e^(−iα) )/2) sin 3x =((e^(3ix) −e^(−3ix) )/(2i))=((e^(6ix) −1)/(2ie^(3ix) ))= =(((e^(3ix) −1)(e^(3ix) +1))/(2ie^(3ix) ))= =((e^(i((3x)/2)) (e^(i((3x)/2)) −e^(−i((3x)/2)) )e^(i((3x)/2)) (e^(i((3x)/2)) +e^(−i((3x)/2)) ))/(2ie^(3ix) ))= =(((e^(i((3x)/2)) −e^(−i((3x)/2)) )(e^(i((3x)/2)) +e^(−i((3x)/2)) ))/(2i))= =2(((e^(i((3x)/2)) −e^(−i((3x)/2)) )(e^(i((3x)/2)) +e^(−i((3x)/2)) ))/(4i))= =2(((e^(i((3x)/2)) −e^(−i((3x)/2)) ))/(2i))×(((e^(i((3x)/2)) +e^(−i((3x)/2)) ))/2)= =2sin ((3x)/2) cos ((3x)/2)

$$\mathrm{sin}\:\alpha\:=\frac{\mathrm{e}^{\mathrm{i}\alpha} −\mathrm{e}^{−\mathrm{i}\alpha} }{\mathrm{2i}}\wedge\mathrm{cos}\:\alpha\:=\frac{\mathrm{e}^{\mathrm{i}\alpha} +\mathrm{e}^{−\mathrm{i}\alpha} }{\mathrm{2}} \\ $$$$\mathrm{sin}\:\mathrm{3}{x}\:=\frac{\mathrm{e}^{\mathrm{3i}{x}} −\mathrm{e}^{−\mathrm{3i}{x}} }{\mathrm{2i}}=\frac{\mathrm{e}^{\mathrm{6i}{x}} −\mathrm{1}}{\mathrm{2ie}^{\mathrm{3i}{x}} }= \\ $$$$=\frac{\left(\mathrm{e}^{\mathrm{3i}{x}} −\mathrm{1}\right)\left(\mathrm{e}^{\mathrm{3i}{x}} +\mathrm{1}\right)}{\mathrm{2ie}^{\mathrm{3i}{x}} }= \\ $$$$=\frac{\mathrm{e}^{\mathrm{i}\frac{\mathrm{3}{x}}{\mathrm{2}}} \left(\mathrm{e}^{\mathrm{i}\frac{\mathrm{3}{x}}{\mathrm{2}}} −\mathrm{e}^{−\mathrm{i}\frac{\mathrm{3}{x}}{\mathrm{2}}} \right)\mathrm{e}^{\mathrm{i}\frac{\mathrm{3}{x}}{\mathrm{2}}} \left(\mathrm{e}^{\mathrm{i}\frac{\mathrm{3}{x}}{\mathrm{2}}} +\mathrm{e}^{−\mathrm{i}\frac{\mathrm{3}{x}}{\mathrm{2}}} \right)}{\mathrm{2ie}^{\mathrm{3i}{x}} }= \\ $$$$=\frac{\left(\mathrm{e}^{\mathrm{i}\frac{\mathrm{3}{x}}{\mathrm{2}}} −\mathrm{e}^{−\mathrm{i}\frac{\mathrm{3}{x}}{\mathrm{2}}} \right)\left(\mathrm{e}^{\mathrm{i}\frac{\mathrm{3}{x}}{\mathrm{2}}} +\mathrm{e}^{−\mathrm{i}\frac{\mathrm{3}{x}}{\mathrm{2}}} \right)}{\mathrm{2i}}= \\ $$$$=\mathrm{2}\frac{\left(\mathrm{e}^{\mathrm{i}\frac{\mathrm{3}{x}}{\mathrm{2}}} −\mathrm{e}^{−\mathrm{i}\frac{\mathrm{3}{x}}{\mathrm{2}}} \right)\left(\mathrm{e}^{\mathrm{i}\frac{\mathrm{3}{x}}{\mathrm{2}}} +\mathrm{e}^{−\mathrm{i}\frac{\mathrm{3}{x}}{\mathrm{2}}} \right)}{\mathrm{4i}}= \\ $$$$=\mathrm{2}\frac{\left(\mathrm{e}^{\mathrm{i}\frac{\mathrm{3}{x}}{\mathrm{2}}} −\mathrm{e}^{−\mathrm{i}\frac{\mathrm{3}{x}}{\mathrm{2}}} \right)}{\mathrm{2i}}×\frac{\left(\mathrm{e}^{\mathrm{i}\frac{\mathrm{3}{x}}{\mathrm{2}}} +\mathrm{e}^{−\mathrm{i}\frac{\mathrm{3}{x}}{\mathrm{2}}} \right)}{\mathrm{2}}= \\ $$$$=\mathrm{2sin}\:\frac{\mathrm{3}{x}}{\mathrm{2}}\:\mathrm{cos}\:\frac{\mathrm{3}{x}}{\mathrm{2}} \\ $$

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