Question

If $1^0=\alpha$ radians, then the approximate value of $\cos {60}^{\circ }{1}^{\prime }$ 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. $\frac{1}{2}$

Answer 1

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

$f\left(\theta \right)=\cos \theta$

$\Rightarrow f^{\prime }\left(\theta \right)=-\sin \theta$

$\Rightarrow f^{\prime }\left({60}^{\circ }\right)=-\sin {60}^{\circ }=\frac{-\sqrt{3}}{2}$
We know $f(x+h)=f\left(x\right)+f^{\prime }\left(x\right)△h$

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

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