# Prove that sinh(x+y)=sinxcoshy+coshxsinhy

Question
Prove that $$\displaystyle{\text{sinh}{{\left({x}+{y}\right)}}}={\sin{{x}}}{\text{cosh}{{y}}}+{\text{cosh}{{x}}}{\text{sinh}{{y}}}$$

2021-02-28
$$\displaystyle{\text{sinh}{{\left({A}\right)}}}=\frac{{{e}ˣ-{e}⁻ˣ}}{{2}}$$ by definition when A=x, so, if A=x+y:
$$\displaystyle{\text{sinh}{{\left({x}+{y}\right)}}}=\frac{{{e}^{{{x}+{y}}}-{\left({e}^{{-{{\left({x}+{y}\right)}}}}\right)}}}{{2}}.$$
$$\displaystyle{\text{cosh}{{\left({x}\right)}}}=\frac{{{e}ˣ+{e}⁻ˣ}}{{2}}$$ by definition.
$$\displaystyle{\text{sinh}{{\left({x}\right)}}}{\text{cosh}{{\left({y}\right)}}}={\left({e}^{{x}}-{e}^{{-{{x}}}}\right)}\frac{{{e}^{{y}}+{e}^{{-{{y}}}}}}{{4}}=\frac{{{e}^{{{x}+{y}}}+{e}^{{{x}-{y}}}-{e}^{{{y}-{x}}}-{e}^{{-{{\left({x}+{y}\right)}}}}}}{{4}}.$$
$$\displaystyle{\text{cosh}{{\left({x}\right)}}}{\text{sinh}{{\left({y}\right)}}}={\left({e}^{{x}}+{e}^{{-{{x}}}}\right)}\frac{{{e}^{{y}}-{e}^{{-{{y}}}}}}{{4}}=\frac{{{e}^{{{x}+{y}}}-{e}^{{{x}-{y}}}+{e}^{{{y}-{x}}}-{e}^{{-{{\left({x}+{y}\right)}}}}}}{{4}}.$$
When we add these last two equations we get sinh(x)cosh(y)+cosh(x)sinh(y)=ZSK
$$\displaystyle{2}\frac{{{e}^{{{x}+{y}}}-{e}^{{-{{\left({x}+{y}\right)}}}}}}{{4}}=\frac{{{e}^{{{x}+{y}}}-{e}^{{-{{\left({x}+{y}\right)}}}}}}{{2}}={\text{sinh}{{\left({x}+{y}\right)}}}$$

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