Question Number 22102 by Parvez akhter last updated on 11/Oct/17
$$\mathrm{cosh}\:\left(\mathrm{15}\right) \\ $$
Answered by $@ty@m last updated on 11/Oct/17
$$\mathrm{cos}\:{hx}=\frac{{e}^{{x}} +{e}^{−{x}} }{\mathrm{2}} \\ $$$$\mathrm{cos}\:{h}\left(\mathrm{15}\right)=\frac{{e}^{\mathrm{15}} +{e}^{−\mathrm{15}} }{\mathrm{2}} \\ $$$$=\frac{{e}^{\mathrm{15}} +\frac{\mathrm{1}}{{e}^{\mathrm{15}} }}{\mathrm{2}} \\ $$
Commented by $@ty@m last updated on 11/Oct/17
