Question

Using differentials, find the approximate values of the following:
(xi) $\cos {61}^{\circ }$, it being given that $\sin {60}^{\circ }=0.86603\text{and}{1}^{\circ }=0.01745radian$.

Answer 1

Let $y=\cos x\Rightarrow \frac{dy}{dx}=-\sin x$
And $x={60}^{\circ }$ and $\Delta x=1^{\circ }$
Then,
$\Delta y=\cos (x+\Delta x)-\cos x$
$\Rightarrow \Delta y=\cos ({60}^{\circ }+1^{\circ })-\cos {60}^{\circ }$
$\Rightarrow \Delta y=\cos {61}^{\circ }-0.5$
$\Rightarrow \cos {61}^{\circ }=0.5+\Delta y$ ...(1)
Now, approximate change in value of y is given by,
$\because \Delta y\approx \left(\frac{dy}{dx}\right)\times \Delta x$
$\Rightarrow \Delta y\approx -\sin x\times \Delta x$
$\Rightarrow \Delta y\approx -\sin {60}^{\circ }\times 1^{\circ }$
$\Rightarrow \Delta y\approx -0.86603\times 0.01745\approx -0.0151$
From equation (1)
$\cos {61}^{\circ }\approx 0.5-0.0151$
$\therefore \cos {61}^{\circ }\approx 0.4849$