Question
Using differentials, find the approximate values of the following:
$\cos\Big(\frac{11\pi}{36}\Big)$

Answer

Consider the function $\text{y}=\text{f} (\text{x})=\cos\text{x}$ Let:
$\text{x}=\frac{\pi}{3}$ $\text{x}+\triangle \text{x}=\frac{11\pi}{36}$ Then,
$\triangle\text{x}= \frac{-\pi}{36}=-5^\circ$ For $\text{x}=\frac{\pi} {3}$
$\text{y}=\cos\Big (\frac{\pi}{3}\Big)=0.5$ Let:
$\text{dx}=\triangle \text{x}=-\sin5^\circ=-0.08716$ Now $\text{y}=\cos\text {x}$
$\Rightarrow\frac {\text{dy}}{\text{dx}}=-\sin\text{x}$ $\Rightarrow\Big(\frac {\text{dy}}{\text{dx}}\Big)_{\text{x}= \frac{\pi}{3}}=0.86603$ $\therefore\triangle \text{y}=\text{dy}=\frac{\text{dy}} {\text{dx}}\text{dx}=-0.86603\times(- 0.08716)=0.075575$ $\Rightarrow\triangle \text{y} =0.075575$ $\therefore\cos\frac {11\pi}{36}=\text{y}+\triangle\text{y} =0.5+0.075570=0.575575$

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