# How do you use differentials to estimate the value of cos(63)?

Mar 13, 2015

Approximating by differential = linear approximation = use the tangent line at a nearby point.

I will assume that you want to approximate $\cos \left({63}^{\circ}\right)$ rather than cosine of 63 radians (or cosine of the number 63).

At some point, either before we begin or within the functions we use, we'll have to change to radian measure. (Really we change to numbers.

Taking $y = \cos x$, we approximate:

$\cos \left({63}^{\circ}\right) \approx \cos \left({60}^{\circ}\right) + \mathrm{dy}$

$\mathrm{dy} = - \sin \left(\frac{\pi}{3}\right) \mathrm{dx} = - \frac{\sqrt{3}}{2} \mathrm{dx}$

$\mathrm{dx} = {63}^{\circ} - {60}^{\circ} = {3}^{\circ} = 3 \left(\frac{\pi}{180}\right) = \frac{\pi}{60}$
(You could call the last 2 "radians", but I won't.)

So $\mathrm{dy} = - \frac{\sqrt{3}}{2} \frac{\pi}{60} = - \frac{\pi \sqrt{3}}{120}$

And $\cos \left({63}^{\circ}\right) \approx \cos \left({60}^{\circ}\right) + \mathrm{dy}$

$\cos \left({63}^{\circ}\right) \approx \frac{1}{2} + \left(- \frac{\pi \sqrt{3}}{120}\right)$

More General:
I find the ideas more clear if we think about linear approximation, which is just approximating by a tangent line.
We have $y = f \left(x\right)$, and we note that
$f \left(x\right) = f \left(a\right) + \Delta y$.

The linear (differential) approximation of $f \left(x\right)$ near $a$ is:
$f \left(x\right) \approx f \left(a\right) + f ' \left(a\right) \left(x - a\right)$, or

$f \left(x\right) \approx f \left(a\right) + f ' \left(a\right) \mathrm{dx} = f \left(a\right) + \mathrm{dy}$

(This is the equation of the line tangent to the graph of the function at the point $\left(a , f \left(a\right)\right)$.)

We are approximating $\Delta y$ by using the change along the tangent (which is $\mathrm{dy}$) to approximate the change on the graph of the function (which is $\Delta y$).