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. $x$
$\frac{dy}{dx}=-\sin (x)$
$\Rightarrow {\left(\frac{dy}{dx}\right)}_{x=\frac{\pi }{6}}=-\sin \frac{\pi }{6}$
$\Rightarrow {\left(\frac{dy}{dx}\right)}_{x=\frac{\pi }{6}}=-\frac{1}{2}$
Let $\Delta y$ be the change in $y$ due to the change $\Delta x$ in $x$.
$\therefore \Delta y=\frac{dy}{dx}\times \Delta x$
$=\left(-\frac{1}{2}\right)\times 0.0174$
$=(-0.5)\times 0.0174\approx -0.0087$
$\therefore f\left(36^{\circ }\right)=y+\Delta y$
$=\frac{\sqrt{3}}{2}-0.0087$
$=\frac{1.732}{2}-0.0087$
$=0.8660-0.0087$
$=0.8573$