Question

The approximate value of $\cos 31^{\circ}$ is (Take $1^{\circ}=0.0174$ )

• A. 0.7521
• B. 0.866
• C. 0.7146
• D. 0.8573

Answer 1

Answer: (D)

Let $y=\cos (x), x=30^{\circ}$ and $x+\Delta x=31^{\circ}$
Let us find $(y)_{x=\frac{\pi}{6}}=\cos \left(\frac{\pi}{6}\right)=\frac{\sqrt{3}}{2}$
Then, $\Delta x=1^{\circ}=0 . 0174$ radian.
Consider the given function,
$y=f(x)=\cos (x)$
Differentiating w.r.t. $x$
$ \frac{d y}{d x} =-\sin (x) $
$ \Rightarrow \left(\frac{d y}{d x}\right)_{x=\frac{\pi}{6}} =-\sin \frac{\pi}{6} $
$\Rightarrow \left(\frac{d y}{d x}\right)_{x=\frac{\pi}{6}} =-\frac{1}{2}$
Let $\Delta y$ be the change in $y$ due to the change $\Delta x$ in $x$.
$\therefore \Delta y=\frac{d y}{d x} \times \Delta x$
$ =\left(-\frac{1}{2}\right) \times 0 .0174 $
$ =(-0.5) \times 0 . 0174 \approx-0 . 0087 $
$ \therefore f\left(36^{\circ}\right) =y+\Delta y $
$ = \frac{\sqrt{3}}{2}-0 .0087 $
$= \frac{1 . 732}{2}-0 .0087$
$ = 0 .8660-0 .0087 $
$ =0 . 8573$