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Question Number 86737 by M±th+et£s last updated on 30/Mar/20

prove that 1/cos2x+cosx+1=((sin((5x)/2))/(2sin(x/2)))+(1/2) 2/((cos(x)+isin(x)−1)/(cos(x)+isin(x)+1))=−i tan(x) 3/((cos(5x)+isin(5x)+1)/(cos(5x)−isin(x)+1))=cos(5x)+isin(5x)

$${prove}\:{that} \\ $$$$\mathrm{1}/{cos}\mathrm{2}{x}+{cosx}+\mathrm{1}=\frac{{sin}\frac{\mathrm{5}{x}}{\mathrm{2}}}{\mathrm{2}{sin}\frac{{x}}{\mathrm{2}}}+\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{2}/\frac{{cos}\left({x}\right)+{isin}\left({x}\right)−\mathrm{1}}{{cos}\left({x}\right)+{isin}\left({x}\right)+\mathrm{1}}=−{i}\:{tan}\left({x}\right) \\ $$$$ \\ $$$$\mathrm{3}/\frac{{cos}\left(\mathrm{5}{x}\right)+{isin}\left(\mathrm{5}{x}\right)+\mathrm{1}}{{cos}\left(\mathrm{5}{x}\right)−{isin}\left({x}\right)+\mathrm{1}}={cos}\left(\mathrm{5}{x}\right)+{isin}\left(\mathrm{5}{x}\right) \\ $$

Commented by som(math1967) last updated on 31/Mar/20

2/(((cosx+isinx−1)(cosx−isinx+1))/((cosx+isinx+1)(cosx−isinx+1))) =((cos^2 x−(1−isinx)^2 )/((cosx+1)^2 −(isinx)^2 )) =((cos^2 x+sin^2 x−1+2isinx)/(cos^2 x+sin^2 x+1+2cosx)) =((2isinx)/(2(1+cosx)))=((2isin(x/2)cos(x/2))/(2cos^2 (x/2))) =itan(x/2) ?????

$$\mathrm{2}/\frac{\left({cosx}+{isinx}−\mathrm{1}\right)\left({cosx}−{isinx}+\mathrm{1}\right)}{\left({cosx}+{isinx}+\mathrm{1}\right)\left({cosx}−{isinx}+\mathrm{1}\right)} \\ $$$$=\frac{{cos}^{\mathrm{2}} {x}−\left(\mathrm{1}−{isinx}\right)^{\mathrm{2}} }{\left({cosx}+\mathrm{1}\right)^{\mathrm{2}} −\left({isinx}\right)^{\mathrm{2}} } \\ $$$$=\frac{{cos}^{\mathrm{2}} {x}+{sin}^{\mathrm{2}} {x}−\mathrm{1}+\mathrm{2}{isinx}}{{cos}^{\mathrm{2}} {x}+{sin}^{\mathrm{2}} {x}+\mathrm{1}+\mathrm{2}{cosx}} \\ $$$$=\frac{\mathrm{2}{isinx}}{\mathrm{2}\left(\mathrm{1}+{cosx}\right)}=\frac{\mathrm{2}{isin}\frac{{x}}{\mathrm{2}}{cos}\frac{{x}}{\mathrm{2}}}{\mathrm{2}{cos}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}} \\ $$$$={itan}\frac{{x}}{\mathrm{2}}\:????? \\ $$

Commented by Prithwish Sen 1 last updated on 31/Mar/20

2. ((e^(ix) −1)/(e^(ix) +1)) = (((e^(i(x/2)) −e^(−i(x/2)) )/2)/(i(((e^(i(x/2)) +e^(−i(x/2)) )/(2i))))) = ((sin(x/2))/(icos(x/2))) = −itan(x/2) please check.

$$\mathrm{2}.\:\:\:\:\:\frac{\mathrm{e}^{\mathrm{ix}} −\mathrm{1}}{\mathrm{e}^{\mathrm{ix}} +\mathrm{1}}\:=\:\frac{\frac{\mathrm{e}^{\mathrm{i}\frac{\mathrm{x}}{\mathrm{2}}} −\mathrm{e}^{−\mathrm{i}\frac{\mathrm{x}}{\mathrm{2}}} }{\mathrm{2}}}{\mathrm{i}\left(\frac{\mathrm{e}^{\mathrm{i}\frac{\mathrm{x}}{\mathrm{2}}} +\mathrm{e}^{−\mathrm{i}\frac{\mathrm{x}}{\mathrm{2}}} }{\mathrm{2i}}\right)}\:=\:\frac{\mathrm{sin}\frac{\boldsymbol{\mathrm{x}}}{\mathrm{2}}}{\mathrm{icos}\frac{\boldsymbol{\mathrm{x}}}{\mathrm{2}}}\:=\:−\mathrm{itan}\frac{\boldsymbol{\mathrm{x}}}{\mathrm{2}} \\ $$$$\boldsymbol{\mathrm{please}}\:\boldsymbol{\mathrm{check}}. \\ $$

Commented by M±th+et£s last updated on 31/Mar/20

yes sir its (x/2)

$${yes}\:{sir}\:{its}\:\frac{{x}}{\mathrm{2}} \\ $$

Commented by M±th+et£s last updated on 31/Mar/20

you are right

$${you}\:{are}\:{right} \\ $$

Answered by jagoll last updated on 31/Mar/20

Commented by M±th+et£s last updated on 31/Mar/20

thank you sir

$${thank}\:{you}\:{sir} \\ $$

Answered by som(math1967) last updated on 31/Mar/20

3/(({cos(5x)+isin(5x)+1}^2 )/({cos(5x)+1}^2 −{isin(5x)}^2 )) =((cos^2 (5x)−sin^2 (5x)+1+2cos(5x)+2isin(5x)+2isin(5x)cos(5x))/(cos^2 (5x)+sin^2 (5x)+1+2cos(5x))) [∵ i^2 =−1] =((2cos^2 (5x)+2cos(5x)+2isin(5x){1+cos(5x)})/(2{1+cos(5x)})) =((2cos(5x){1+cos(5x)}+2isin(5x){1+cos(5x)})/(2{1+cos(5x)})) =((2{1+cos(5x)}{cos(5x)+isin(5x)})/(2{1+cos(5x)})) =cos(5x)+isin(5x)

$$\mathrm{3}/\frac{\left\{{cos}\left(\mathrm{5}{x}\right)+{isin}\left(\mathrm{5}{x}\right)+\mathrm{1}\right\}^{\mathrm{2}} }{\left\{{cos}\left(\mathrm{5}{x}\right)+\mathrm{1}\right\}^{\mathrm{2}} −\left\{{isin}\left(\mathrm{5}{x}\right)\right\}^{\mathrm{2}} } \\ $$$$=\frac{{cos}^{\mathrm{2}} \left(\mathrm{5}{x}\right)−{sin}^{\mathrm{2}} \left(\mathrm{5}{x}\right)+\mathrm{1}+\mathrm{2}{cos}\left(\mathrm{5}{x}\right)+\mathrm{2}{isin}\left(\mathrm{5}{x}\right)+\mathrm{2}{isin}\left(\mathrm{5}{x}\right){cos}\left(\mathrm{5}{x}\right)}{{cos}^{\mathrm{2}} \left(\mathrm{5}{x}\right)+{sin}^{\mathrm{2}} \left(\mathrm{5}{x}\right)+\mathrm{1}+\mathrm{2}{cos}\left(\mathrm{5}{x}\right)} \\ $$$$\left[\because\:{i}^{\mathrm{2}} =−\mathrm{1}\right] \\ $$$$=\frac{\mathrm{2}{cos}^{\mathrm{2}} \left(\mathrm{5}{x}\right)+\mathrm{2}{cos}\left(\mathrm{5}{x}\right)+\mathrm{2}{isin}\left(\mathrm{5}{x}\right)\left\{\mathrm{1}+{cos}\left(\mathrm{5}{x}\right)\right\}}{\mathrm{2}\left\{\mathrm{1}+{cos}\left(\mathrm{5}{x}\right)\right\}} \\ $$$$=\frac{\mathrm{2}{cos}\left(\mathrm{5}{x}\right)\left\{\mathrm{1}+{cos}\left(\mathrm{5}{x}\right)\right\}+\mathrm{2}{isin}\left(\mathrm{5}{x}\right)\left\{\mathrm{1}+{cos}\left(\mathrm{5}{x}\right)\right\}}{\mathrm{2}\left\{\mathrm{1}+{cos}\left(\mathrm{5}{x}\right)\right\}} \\ $$$$=\frac{\mathrm{2}\left\{\mathrm{1}+{cos}\left(\mathrm{5}{x}\right)\right\}\left\{{cos}\left(\mathrm{5}{x}\right)+{isin}\left(\mathrm{5}{x}\right)\right\}}{\mathrm{2}\left\{\mathrm{1}+{cos}\left(\mathrm{5}{x}\right)\right\}} \\ $$$$={cos}\left(\mathrm{5}{x}\right)+{isin}\left(\mathrm{5}{x}\right) \\ $$

Commented by M±th+et£s last updated on 31/Mar/20

thank you sir

$${thank}\:{you}\:{sir} \\ $$

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