Question Number 85620 by Roland Mbunwe last updated on 23/Mar/20 | ||
$${prove}\:{that} \\ $$$$ \\ $$$$\mathrm{cosh}\:\left({x}−{y}\right)=\mathrm{cosh}\:{x}\mathrm{cosh}\:{y}−\mathrm{sinh}\:{x}\mathrm{sinh}\:{y} \\ $$ | ||
Answered by Rio Michael last updated on 23/Mar/20 | ||
$$\:{RHS}\:=\:\mathrm{cosh}\:{x}\:\mathrm{cosh}\:{y}\:−\:\mathrm{sinh}\:{x}\:\mathrm{sinh}\:{y} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\:\frac{\mathrm{1}}{\mathrm{2}}\left({e}^{{x}} \:+\:{e}^{−{x}} \right)\:\frac{\mathrm{1}}{\mathrm{2}}\left({e}^{{y}} \:+\:{e}^{−{y}} \right)\:−\:\frac{\mathrm{1}}{\mathrm{2}}\left({e}^{{x}} \:−{e}^{−{x}} \right)\frac{\mathrm{1}}{\mathrm{2}}\left({e}^{{y}} −{e}^{−{y}} \right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\:\frac{\mathrm{1}}{\mathrm{4}}\left[\left({e}^{{x}} +{e}^{−{x}} \right)\left({e}^{{y}} \:+\:{e}^{−{y}} \right)−\left({e}^{{x}} −{e}^{−{x}} \right)\left({e}^{{y}} −{e}^{−{y}} \right)\right] \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\:\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{2}{e}^{\left({x}−{y}\right)} +\:\mathrm{2}{e}^{−\left({x}−{y}\right)} \right)\:=\:\frac{\mathrm{1}}{\mathrm{2}}\left({e}^{\left({x}−{y}\right)} \:+\:{e}^{−\left({x}−{y}\right)} \right)\:=\:\mathrm{cosh}\left({x}−{y}\right) \\ $$ | ||