Q109 / 160 · AP EAMCET 2018 (22 Apr Shift 1) · Mathematics
The approximate value of $\cos 31^{\circ}$ is $\left(\right.$ Take $\left.1^{\circ}=0.0174\right)$
- A0.7521
- B0.866
- C0.7146
- Correct answerD0.8573
Solution
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. $x$,
$$
\begin{aligned}
\frac{d y}{d x} & =-\sin (x) \\
\Rightarrow \quad\left(\frac{d y}{d x}\right)_{x=\frac{\pi}{6}} & =-\sin \frac{\pi}{6} \\
\Rightarrow \quad\left(\frac{d y}{d x}\right)_x=\frac{\pi}{6} & =-\frac{1}{2}
\end{aligned}
$$
Let $\Delta y$ be the change in $y$ due to the change $\Delta x$ in $x$.
$$
\begin{aligned}
\therefore \quad \Delta y & =\frac{d y}{d x} \times \Delta x \\
& =\left(-\frac{1}{2}\right) \times 0.0174 \\
& =(-0.5) \times 0.0174=-0.0087 \\
\therefore \quad f\left(36^{\circ}\right) & =y+\Delta y \\
= & \frac{\sqrt{3}}{2}-0.0087 \\
& =\frac{1.732}{2}-0.0087 \\
& =0.8660-0.0087 \\
& =0.8573
\end{aligned}
$$
Hence, the correct answer is option: 'D'.