Question

If 1∘=α radians then the approximate value of cos(60∘1′) is

Answer 1

Final Answer: $[ \frac{1}{2}- \frac{ \alpha \sqrt{3} }{120} ]$ 

Formula:
1) cos(A+B)=cos(A)cos(B)-sin(A)sin(B)
2) When angle in cos(A) is very small with respect to 1 ,then cos(A) =1 & When angle in sin(A) is very small with respect to 1,then
sin(A) = A.

Steps:
1) We have, 
 $cos(60\:\degree1')=cos(60\degree+1')$

[tex]cos(60\degree+1')=cos(60\degree+ (\frac{1}{60})\degree) \\ \\ = \ \textgreater \ cos(60\degree)cos((\frac{1}{60})\degree)-sin(60\degree)sin((\frac{1}{60})\degree)) \\ \\ =\ \textgreater \ \frac{1}{2}cos((\frac{1}{60})\degree) - \frac{ \sqrt{3} }{2}sin((\frac{1}{60})\degree)) \\ \\ [/tex]

2) Here , we will use 2) Logic: 
Then, 
[tex]LHS= \frac{1}{2}*1- \frac{ \sqrt{3} }{2}*sin( \frac{ \alpha }{60}) \\ \\ = \frac{1}{2} - \frac{ \sqrt{3} }{2}* \frac{ \alpha }{60} \\ \\ = \frac{1}{2}- \frac{ \sqrt{3} \alpha }{120} [/tex]


3) Hence,
Finally , Approximate value of $cos(60\degree\:1')$ is 

$\boxed{\frac{1}{2}- \frac{ \sqrt{3} \alpha }{120}}$