Question

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

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

2021-05-03
Substitute the given function: $$\displaystyle\frac{{{f{{\left({x}+{h}\right)}}}-{f{{\left({x}\right)}}}}}{{h}}=\frac{{{\sin{{\left({x}+{h}\right)}}}-{\sin{{x}}}}}{{h}}$$
Use the the sine of a sum identity: $$\displaystyle{\sin{{\left({A}+{B}\right)}}}={\sin{{A}}}{\cos{{B}}}+{\cos{{A}}}{\sin{{B}}}$$
$$\displaystyle\frac{{{f{{\left({x}+{h}\right)}}}-{f{{\left({x}\right)}}}}}{{h}}=\frac{{{\sin{{x}}}{\text{cosh}{+}}{\cos{{x}}}{\text{sinh}{-}}{\sin{{x}}}}}{{h}}$$
Regroup as: $$\displaystyle\frac{{{f{{\left({x}+{h}\right)}}}-{f{{\left({x}\right)}}}}}{{h}}=\frac{{{\sin{{x}}}{\text{cosh}{-}}{\sin{{x}}}+{\cos{{x}}}{\text{sinh}}}}{{h}}$$
Separate as: $$\displaystyle\frac{{{f{{\left({x}+{h}\right)}}}-{f{{\left({x}\right)}}}}}{{h}}={\left(\frac{{{\sin{{x}}}{\text{cosh}{-}}{\sin{{x}}}}}{{h}}\right)}+{\left(\frac{{{\cos{{x}}}{\text{sinh}}}}{{h}}\right)}$$
Factor out sinxsinx from the first term and cosxcosx from the second term:
$$\displaystyle\frac{{{f{{\left({x}+{h}\right)}}}-{f{{\left({x}\right)}}}}}{{h}}={\sin{{x}}}{\left(\frac{{{\text{cosh}{-}}{1}}}{{h}}\right)}+{\cos{{x}}}{\left(\frac{{\text{sinh}}}{{h}}\right)}$$