Question

If $1^o=\alpha$ radians, then the approximate value of $\cos \left({60}^o{1}^{\prime }\right)$ 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. $\frac{1}{2}+\frac{\alpha }{120}$

Answer 1

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

Using $\cos (A+B)=\cos \left(A\right)\cos \left(B\right)-\sin \left(A\right)\sin \left(B\right)$, we get

$\cos \left({60}^o{1}^{\prime }\right)=\cos \left({60}^o\right)\cos \left(1^{\prime }\right)-\sin \left({60}^o\right)\sin \left(1^{\prime }\right)$

$=\frac{1}{2}-\frac{\sqrt{3}}{2}\times \frac{\alpha }{60}$ ....When $\theta$ is small and in radian, $\sin \theta =\theta$ and $\cos \theta =1$