Find the approximate value of cos(11π36), assuming the value of cos(π3).
Let y=cosx ⋯ (1)
Let x=π3, Δx=−π36 so that
(x+Δx)=π3−π36=11π36 ⋯ (2)
Now Δy=cos(x+Δx)−cosx
i.e. Δy=cos(11π36)−cos(π3)
i.e. cos(11π36)=Δy+cos(π3)
i.e cos(11π36)=Δy+12 ⋯ (3)
Differentiating equation (1) w.r.t x gives
dydx=−sinx=−sin(π3)=−√32 ⋯ (4)
Now Δy is approximately equal to dy and
dy=(dydx)×Δx=−√32×(−π36)=π√372 [from equations (2) and (4)]
∴ \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}