Using Differentials to Approximate Tan 61
In this tutorial we shall look at the use differentials to approximate the value of $$\tan {61^ \circ }$$.
The nearest number to 44 whose tangent value can be taken is 61, so let us consider that $$x = {60^ \circ }$$ and $$\delta x = dx = {1^ \circ }$$.
Now consider
\[y = \tan x\,\,\,\,\,\,{\text{ – – – }}\left( {\text{i}} \right)\]
\[\begin{gathered} y + \delta y = \tan \left( {x + \delta x} \right) \\ \Rightarrow \tan \left( {x + \delta x} \right) = y + \delta y\,\,\,\,{\text{ – – – }}\left( {{\text{ii}}} \right) \\ \end{gathered} \]
Since $$y = \tan x,\,\,\delta y \approx dy$$, putting these values in equation (ii), we have
\[\tan \left( {x + \delta x} \right) \approx \tan x + \delta y\,\,\,\,{\text{ – – – }}\left( {{\text{iii}}} \right)\]
Taking the differential of equation (i), we have
\[dy = d\left( {\tan x} \right) = {\sec ^2}xdx\]
Putting this value in equation (ii), we have
\[\begin{gathered} \tan \left( {x + \delta x} \right) \approx \tan x + {\sec ^2}xdx \\ \Rightarrow \tan \left( {{{60}^ \circ } + {1^ \circ }} \right) \approx \tan {60^ \circ } + {\sec ^2}{60^ \circ }\left( {{1^ \circ }} \right)\,\,\,\,\,\because x = {60^ \circ },\,\,dx = {1^ \circ } = \delta x \\ \Rightarrow \tan {61^ \circ } \approx 1.732 + 4\left( {0.0174} \right)\,\,\,\,\,\because {1^ \circ } = 0.0174 \\ \Rightarrow \tan {61^ \circ } \approx 1.732 + 0.0696 \\ \Rightarrow \tan {61^ \circ } \approx 1.802 \\ \end{gathered} \]
Pius
September 22 @ 8:47 pm
I haven’t understood where 1 degrees changes to 0.0174