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Question Number 112 by saidinesh999.sd@gmail.com last updated on 25/Jan/15

if sinhx=1/2 then find cosh (2x)+sinh(2x)

$${if}\:\:{sinhx}=\mathrm{1}/\mathrm{2}\:\:{then}\:{find}\:{cosh} \\ $$$$\left(\mathrm{2}{x}\right)+{sinh}\left(\mathrm{2}{x}\right) \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

Answered by sagarwal last updated on 04/Dec/14

sinh x=((e^x −e^(−x) )/2) cosh x=((e^x +e^(−x) )/2) cosh 2x + sinh 2x=e^(2x) Let e^x =y sinh x=(1/2)(y−(1/y))=(1/2) y−(1/y)=1 y^2 −y−1=0 y=((1±(√(1+4)))/2)=((1±(√5))/2) Assuming x ∈R, y=e^x so we can discard −ve solution y=((1+(√5))/2) e^(2x) =y^2 =(((1+(√5))/2))^2 =((6+2(√5))/4) =((3+(√5))/2)

$$\mathrm{sinh}\:{x}=\frac{{e}^{{x}} −{e}^{−{x}} }{\mathrm{2}} \\ $$$$\mathrm{cosh}\:{x}=\frac{{e}^{{x}} +{e}^{−{x}} }{\mathrm{2}} \\ $$$$\mathrm{cosh}\:\mathrm{2}{x}\:+\:\mathrm{sinh}\:\mathrm{2}{x}={e}^{\mathrm{2}{x}} \\ $$$$\mathrm{Let}\:{e}^{{x}} ={y} \\ $$$$\mathrm{sinh}\:{x}=\frac{\mathrm{1}}{\mathrm{2}}\left({y}−\frac{\mathrm{1}}{{y}}\right)=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${y}−\frac{\mathrm{1}}{{y}}=\mathrm{1} \\ $$$${y}^{\mathrm{2}} −{y}−\mathrm{1}=\mathrm{0} \\ $$$${y}=\frac{\mathrm{1}\pm\sqrt{\mathrm{1}+\mathrm{4}}}{\mathrm{2}}=\frac{\mathrm{1}\pm\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$$\mathrm{Assuming}\:{x}\:\in\mathbb{R},\:{y}={e}^{{x}} \:\mathrm{so}\:\mathrm{we}\:\mathrm{can} \\ $$$$\mathrm{discard}\:−\mathrm{ve}\:\mathrm{solution} \\ $$$${y}=\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$${e}^{\mathrm{2}{x}} ={y}^{\mathrm{2}} =\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{\mathrm{2}} =\frac{\mathrm{6}+\mathrm{2}\sqrt{\mathrm{5}}}{\mathrm{4}} \\ $$$$=\frac{\mathrm{3}+\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$

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