Question

If $1^0=\alpha$ radians then the approximate value of $\cos {60}^0{1}^{{}^{\prime }}$ is

• A. $\frac{1}{2}+\frac{\alpha \sqrt{3}}{120}$
• B. $\frac{1}{2}-\frac{\alpha }{120}$
• C. $\frac{1}{2}-\frac{\alpha \sqrt{3}}{120}$
• D. none of these

Answer 1

The correct option is C $\frac{1}{2}-\frac{\alpha \sqrt{3}}{120}$

$f\left(x\right)=cos\theta$
$\Rightarrow f^{\prime }\left(x\right)=sin\theta$
$\Rightarrow f^{\prime }\left({60}^{\circ }\right)=sin{60}^{\circ }=\frac{\sqrt{3}}{2}$
$\Rightarrow f\left({60}^{\circ }{1}^{\prime }\right)=sin{60}^{\circ }-\frac{\sqrt{3}}{2}\ast \frac{\alpha }{{60}^{\circ }}$
$\Rightarrow f\left({60}^{\circ }{1}^{\prime }\right)=\frac{1}{2}-\frac{\alpha \sqrt{3}}{{120}^{\circ }}$