Question

Using differentials, find the approximate values of the cos 61°, it being given that sin60° = 0.86603 and 1° = 0.01745 radian ?

Answer 1

\[\text { Consider the function }y = f\left( x \right) = \cos x^\circ . \]

\[\text { Let }: \]

\[ x = 60^\circ \]

\[ x + ∆ x = 61^\circ\]

\[\text { Then }, \]

\[ ∆ x = 1^\circ = 0 . 01745\]

\[\text { For } x = 60^\circ, \]

\[ y = \cos 60^\circ = 0 . 5\]

\[\text { Let }: \]

\[ dx = ∆ x = 0 . 01745\]

\[\text { Now }, y = \cos x\]

\[ \Rightarrow \frac{dy}{dx} = - \sin x\]

\[ \Rightarrow \left( \frac{dy}{dx} \right)_{x = 60} = - 0 . 86603\]

\[ \therefore ∆ y = dy = \frac{dy}{dx}dx = - 0 . 86603 \times 0 . 01745 = - 0 . 01511\]

\[ \Rightarrow ∆ y = - 0 . 01511\]

\[ \therefore \cos 61^\circ = y + ∆ y = 0 . 48488 \approx 0 . 48489\]