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

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

Explanation

Solution

Hint : In order to estimate the value of cos(63)\cos \left( {63} \right) write the value in degree rather than radian. Then use the general formula for approximating the differentials that is f(x)f(a)+f(a)(xa)f\left( x \right) \approx f\left( a \right) + f'\left( a \right)\left( {x - a} \right)or f(x)f(a)+f(a)dx=f(a)+dyf\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(63)\cos \left( {63} \right).
In order to solve it using approximating the differentials use the formula that is f(x)f(a)+f(a)(xa)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(x)=cos63f\left( x \right) = \cos {63^ \circ } that implies x=63x = {63^ \circ }and f(a)=cos60=12f\left( a \right) = \cos {60^ \circ } = \dfrac{1}{2}(We have taken the nearest degree whose value we know) that implies a=60a = {60^ \circ }.
Take f(x)=cosxf\left( x \right) = \cos x in general, that gives f(x)=sinxf'\left( x \right) = - \sin x.
When we put the above value for f(a)f\left( a \right), we get: f(a)=sinaf'\left( a \right) = - \sin a.By putting the value of aa we get:
f(60)=sin60=32f'\left( {{{60}^ \circ }} \right) = - \sin {60^ \circ } = - \dfrac{{\sqrt 3 }}{2}
For, (xa)\left( {x - a} \right), convert the values of xx and aainto radians and put the values and we get:
(xa)=(63π18060π180)=3π180=π60\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(x)f(a)+f(a)(xa)f\left( x \right) \approx f\left( a \right) + f'\left( a \right)\left( {x - a} \right)and we get:

cos(63)cos(60)+(sinπ3)(63π18060π180) cos(63)cos(60)+(sinπ3)(63π18060π180) cos(63)cos(60)+(sinπ3)(63π18060π180) cos(63)12+(32)(π60) cos(63)1232(π60) cos(63)0.50.04536 cos(63)0.45464   \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(63)\cos \left( {63} \right), we get: cos(63)0.45464\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(x)=f(a)+Δyf\left( x \right) = f\left( a \right) + \Delta y . On solving it further we get more solvable results.
f(x)f(a)+f(a)dx=f(a)+dyf\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 (a,f(a))\left( {a,f\left( a \right)} \right).
We approximately find Δy \Delta y by changing the value along the tangent to approximate the change on the graph of the function.