Given that f(x)=cosx, show that (f(x+h)-f(x))/h=cosx((cosh-1)/h)+sinx(sinh/h)

Given that $$\displaystyle{f{{\left({x}\right)}}}={\cos{{x}}}$$, show that $$\displaystyle\frac{{{f{{\left({x}+{h}\right)}}}-{f{{\left({x}\right)}}}}}{{h}}={\cos{{x}}}{\left(\frac{{{\text{cosh}{-}}{1}}}{{h}}\right)}+{\sin{{x}}}{\left(\frac{{\text{sinh}}}{{h}}\right)}$$

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Substitute the given function: $$\displaystyle\frac{{{f{{\left({x}+{h}\right)}}}-{f{{\left({x}\right)}}}}}{{h}}=\frac{{{\cos{{\left({x}+{h}\right)}}}-{\cos{{x}}}}}{{h}}$$
Use the the sine of a sum identity: $$\displaystyle{\sin{{\left({A}+{B}\right)}}}={\sin{{A}}}{\cos{{B}}}+{\cos{{A}}}{\sin{{B}}}$$ $$\displaystyle\frac{{{f{{\left({x}+{h}\right)}}}-{f{{\left({x}\right)}}}}}{{h}}=\frac{{{\cos{{x}}}{\text{cosh}{-}}{\sin{{x}}}{\text{sinh}{-}}{\cos{{x}}}}}{{h}}$$
Regroup as: $$\displaystyle\frac{{{f{{\left({x}+{h}\right)}}}-{f{{\left({x}\right)}}}}}{{h}}=\frac{{{\cos{{x}}}{\text{cosh}{-}}{\cos{{x}}}-{\sin{{x}}}{\text{sinh}}}}{{h}}$$
Separate as: $$\displaystyle\frac{{{f{{\left({x}+{h}\right)}}}-{f{{\left({x}\right)}}}}}{{h}}={\left(\frac{{{\cos{{x}}}{\text{cosh}{-}}{\cos{{x}}}}}{{h}}\right)}-{\left(\frac{{{\sin{{x}}}{\text{sinh}}}}{{h}}\right)}$$
Factor out sinxsinx from the first term and cosxcosx from the second term:
$$\displaystyle\frac{{{f{{\left({x}+{h}\right)}}}-{f{{\left({x}\right)}}}}}{{h}}={\cos{{x}}}{\left(\frac{{{\text{cosh}{-}}{1}}}{{h}}\right)}-{\sin{{x}}}{\left(\frac{{\text{sinh}}}{{h}}\right)}$$
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