Find the approximate value of $\cos \left(\dfrac{11 \pi}{36}\right)$, assuming the value of $\cos \left(\dfrac{\pi}{3}\right)$.
Let $y = \cos x$ $\;\; \cdots$ (1)
Let $x = \dfrac{\pi}{3}$, $\Delta x = -\dfrac{\pi}{36}$ so that
$\left(x + \Delta x\right) = \dfrac{\pi}{3} - \dfrac{\pi}{36} = \dfrac{11\pi}{36}$ $\;\; \cdots$ (2)
Now $\Delta y = \cos \left(x + \Delta x\right) - \cos x$
i.e. $\Delta y = \cos \left(\dfrac{11\pi}{36}\right) - \cos \left(\dfrac{\pi}{3}\right)$
i.e. $\cos \left(\dfrac{11\pi}{36}\right) = \Delta y + \cos \left(\dfrac{\pi}{3}\right)$
i.e $\cos \left(\dfrac{11 \pi}{36}\right) = \Delta y + \dfrac{1}{2}$ $\;\; \cdots$ (3)
Differentiating equation (1) w.r.t x gives
$\dfrac{dy}{dx} = -\sin x = -\sin \left(\dfrac{\pi}{3}\right) = -\dfrac{\sqrt{3}}{2}$ $\;\; \cdots$ (4)
Now $\Delta y$ is approximately equal to dy and
$dy = \left(\dfrac{dy}{dx}\right) \times \Delta x = - \dfrac{\sqrt{3}}{2} \times \left(-\dfrac{\pi}{36}\right) = \dfrac{\pi \sqrt{3}}{72}$ [from equations (2) and (4)]
$\therefore$ $\Delta y \approx dy = \dfrac{\pi \sqrt{3}}{72}$ $\;\; \cdots$ (5)
$\therefore$ We have from equations (3) and (5)
$\cos \left(\dfrac{11 \pi}{36}\right) = \dfrac{\pi \sqrt{3}}{72} + \dfrac{1}{2}$