Question Number 46534 by peter frank last updated on 28/Oct/18
$$\mathrm{using}\:\mathrm{taylors}\:\mathrm{expansion} \\ $$$$\mathrm{find}\:\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of} \\ $$$$\left.\mathrm{a}\right)\mathrm{tan45}°\:\mathrm{1}'\: \\ $$$$\left.\mathrm{b}\right)\mathrm{sin30}°\:\mathrm{1}' \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 28/Oct/18
$$\left.{a}\right){y}={f}\left({x}\right)\:\:{when}\:{x}\:{changes}\:{to}\:{x}+\bigtriangleup{x} \\ $$$${y}\:{changes}\:{to}\:{y}+\bigtriangleup{y} \\ $$$${here}\:{f}\left({x}\right)={tanx}\:\: \\ $$$${y}={tanx}\:\:\frac{{dy}}{{dx}}={sec}^{\mathrm{2}} {x} \\ $$$$\frac{{dy}}{{dx}}=\frac{\bigtriangleup{y}}{\bigtriangleup{x}} \\ $$$$\bigtriangleup{y}=\frac{{dy}}{{dx}}\bigtriangleup{x} \\ $$$$\bigtriangleup{y}={sec}^{\mathrm{2}} {x}.\bigtriangleup{x} \\ $$$$\bigtriangleup{x}=\mathrm{1}^{'} =\left(\frac{\mathrm{1}}{\mathrm{60}}\right)^{{o}} =\frac{\pi}{\mathrm{180}}×\frac{\mathrm{1}}{\mathrm{60}}=\mathrm{0}.\mathrm{00029} \\ $$$$\bigtriangleup{y}={sec}^{\mathrm{2}} \mathrm{45}^{{o}} ×\mathrm{0}.\mathrm{00029}=\mathrm{0}.\mathrm{00058} \\ $$$${so}\:{value}\:{of}\:{tan}\left(\mathrm{45}^{{o}} \mathrm{1}'\right)=\mathrm{1}+\mathrm{0}.\mathrm{00058}=\mathrm{1}.\mathrm{00058} \\ $$$$ \\ $$$$\left.{b}\right)\bigtriangleup{y}=\frac{{dy}}{{dx}}\bigtriangleup{x} \\ $$$${y}={sinx}\:\:\:\frac{{dy}}{{dx}}={cosx} \\ $$$$\bigtriangleup{y}={cosx}.\bigtriangleup{x} \\ $$$$\bigtriangleup{y}={cos}\mathrm{30}^{{o}} ×\mathrm{0}.\mathrm{00029}=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}×\mathrm{0}.\mathrm{00029}=\mathrm{0}.\mathrm{00025} \\ $$$${so}\:{sin}\left(\mathrm{30}^{{o}} \mathrm{1}'\right)=\mathrm{0}.\mathrm{5}+\mathrm{0}.\mathrm{00025}=\mathrm{0}.\mathrm{50025} \\ $$