Question

How do you evaluate `cos 870`?

Answer 1

`cos870^o=-sqrt3/2`

Explanation:

As trigonometric ratios get repeated over a cycle over `2pi` or `360^o`.

Hence `cos870^o=cos(360^o xx2+150^o)=cos150^o`

Further as `cos(180^o-A)=-cosA`

`cos870^o=cos150^o=cos(180^o-30^o)=-cos30^o=-sqrt3/2`