Question

If $1^\circ = \alpha $ radians, then the approximate value of $\cos \left( {60^\circ 1'} \right)$ is
$
  {\text{a}}{\text{. }}\dfrac{1}{2} + \dfrac{{\alpha \sqrt 3 }}{{120}} \\
  {\text{b}}{\text{. }}\dfrac{1}{2} - \dfrac{\alpha }{{120}} \\
  {\text{c}}{\text{. }}\dfrac{1}{2} - \dfrac{{\alpha \sqrt 3 }}{{120}} \\
  {\text{d}}{\text{. }}\dfrac{1}{2} + \dfrac{{\alpha \sqrt 3 }}{{120}} \\
$

Answer 1

Hint: Write $\cos \left( {60^\circ 1'} \right) = \cos \left( {60^\circ + 1'} \right)$. Then use the cos(a+b) formula to solve it.

Complete step-by-step answer:
We have to find out the approximate value of $\cos \left( {60^\circ 1'} \right)$
$\cos \left( {60^\circ 1'} \right) = \cos \left( {60^\circ + 1'} \right)$
As we know $\cos \left( {a + b} \right) = \cos a\cos b - \sin a\sin b$, using it, we get
$\cos \left( {60^\circ 1'} \right) = \cos \left( {60^\circ + 1'} \right) = \cos 60^\circ \cos 1' - \sin 60^\circ \sin 1'..............\left( 1 \right)$
Now as we know 1 minute i.e. $1' = \dfrac{{1^\circ }}{{60^\circ }} = \dfrac{\alpha }{{60^\circ }}$ as $1^\circ = \alpha$ radians (given)
As we know the value of $\cos 60^\circ = \dfrac{1}{2},{\text{ }}\sin 60^\circ = \dfrac{{\sqrt 3 }}{2}$
Substitute these value in equation 1
$
   \Rightarrow \cos \left( {60^\circ 1'} \right) = \cos 60^\circ \cos 1' - \sin 60^\circ \sin 1' \\
   \Rightarrow \cos \left( {60^\circ 1'} \right) = \dfrac{1}{2}\cos \left( {\dfrac{\alpha }{{60^\circ }}} \right) - \dfrac{{\sqrt 3 }}{2}\sin \left( {\dfrac{\alpha }{{60^\circ }}} \right)..........\left( 2 \right) \\
$

Now,
$\left( {\dfrac{\alpha }{{60^\circ }}} \right) < < < < < 1$
Therefore, approximate value of
 $\cos \left( {\dfrac{\alpha }{{60^\circ }}} \right) \simeq 1,{\text{ sin}}\left( {\dfrac{\alpha }{{60^\circ }}} \right) \simeq \dfrac{\alpha }{{60^\circ }}$
Therefore from equation 2
$\cos \left( {60^\circ 1'} \right) = \dfrac{1}{2}\cos \left( {\dfrac{\alpha }{{60^\circ }}} \right) - \dfrac{{\sqrt 3 }}{2}\sin \left( {\dfrac{\alpha }{{60^\circ }}} \right) = \dfrac{1}{2} \times 1 - \dfrac{{\sqrt 3 }}{2} \times \dfrac{\alpha }{{60^\circ }}$
$ \Rightarrow \cos \left( {60^\circ 1'} \right) = \dfrac{1}{2} - \dfrac{{\alpha \sqrt 3 }}{{120^\circ }}$
Hence the approximate value of $\cos \left( {60^\circ 1'} \right) = \dfrac{1}{2} - \dfrac{{\alpha \sqrt 3 }}{{120^\circ }}$
Hence option (c) is correct.

Note: In these types of problems, it is crucial to remember the sine and cosine of sum of angles and also know the approximate value of $\cos a$ and $\sin a$ if a is very small, then after simplification we will get the required approximate value.