How do you use differentials to estimate the value of $\cos \left( {63} \right)$?

Hint: In order to estimate the value of $\cos \left( {63} \right)$ write the value in degree rather than radian. Then use the general formula for approximating the differentials that is $f\left( x \right) \approx f\left( a \right) + f'\left( a \right)\left( {x - a} \right)$or $f\left( x \right) \approx f\left( a \right) + f'\left( a \right)dx = f\left( a \right) + dy$, Just find the values and put them, in the formula and get the value needed.

Complete step by step solution:
We are given the value $\cos \left( {63} \right)$.
In order to solve it using approximating the differentials use the formula that is $f\left( x \right) \approx f\left( a \right) + f'\left( a \right)\left( {x - a} \right)$.Just find the values to fit in the formula and get the results.
Let’s take $f\left( x \right) = \cos {63^ \circ }$ that implies $x = {63^ \circ }$and $f\left( a \right) = \cos {60^ \circ } = \dfrac{1}{2}$(We have taken the nearest degree whose value we know) that implies $a = {60^ \circ }$.
Take $f\left( x \right) = \cos x$ in general, that gives $f'\left( x \right) = - \sin x$.
When we put the above value for $f\left( a \right)$, we get: $f'\left( a \right) = - \sin a$.By putting the value of \[a\] we get:
$f'\left( {{{60}^ \circ }} \right) = - \sin {60^ \circ } = - \dfrac{{\sqrt 3 }}{2}$
For, $\left( {x - a} \right)$, convert the values of $x$ and $a$into radians and put the values and we get:
$\left( {x - a} \right) = \left( {\dfrac{{63\pi }}{{180}} - \dfrac{{60\pi }}{{180}}} \right) = \dfrac{{3\pi }}{{180}} = \dfrac{\pi }{{60}}$.
Now, put the values of each term in the formula $f\left( x \right) \approx f\left( a \right) + f'\left( a \right)\left( {x - a} \right)$and we get:
 \[
  \cos \left( {{{63}^ \circ }} \right) \approx \cos \left( {{{60}^ \circ }} \right) + \left( { - \sin \dfrac{\pi }{3}} \right)\left( {\dfrac{{63\pi }}{{180}} - \dfrac{{60\pi }}{{180}}} \right) \\
  \cos \left( {{{63}^ \circ }} \right) \approx \cos \left( {{{60}^ \circ }} \right) + \left( { - \sin \dfrac{\pi }{3}} \right)\left( {\dfrac{{63\pi }}{{180}} - \dfrac{{60\pi }}{{180}}} \right) \\
  \cos \left( {{{63}^ \circ }} \right) \approx \cos \left( {{{60}^ \circ }} \right) + \left( { - \sin \dfrac{\pi }{3}} \right)\left( {\dfrac{{63\pi }}{{180}} - \dfrac{{60\pi }}{{180}}} \right) \\
  \cos \left( {{{63}^ \circ }} \right) \approx \dfrac{1}{2} + \left( { - \dfrac{{\sqrt 3 }}{2}} \right)\left( {\dfrac{\pi }{{60}}} \right) \\
  \cos \left( {{{63}^ \circ }} \right) \approx \dfrac{1}{2} - \dfrac{{\sqrt 3 }}{2}\left( {\dfrac{\pi }{{60}}} \right) \\
  \cos \left( {{{63}^ \circ }} \right) \approx 0.5 - 0.04536 \\
  \cos \left( {{{63}^ \circ }} \right) \approx 0.45464 \;
 \]

Therefore, by using differentials to estimate the value of $\cos \left( {63} \right)$, we get: \[\cos \left( {{{63}^ \circ }} \right) \approx 0.45464\]
So, the correct answer is “0.45464”.

Note: Approximating by differential is also called linear approximation or using the tangent line at a nearby point.
The most general formula for linear approximation is \[f\left( x \right) = f\left( a \right) + \Delta y\] . On solving it further we get more solvable results.
 $f\left( x \right) \approx f\left( a \right) + f'\left( a \right)dx = f\left( a \right) + dy$ is the equation for a line tangent for the graph of the function at the point $\left( {a,f\left( a \right)} \right)$.
 We approximately find \[ \Delta y\] by changing the value along the tangent to approximate the change on the graph of the function.