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Answer 1 · 2017-12-11T19:13:19

# = sqrt3 /2 #

The first thing we must use and consider is what #cos 30 # is?

We know #cos 30 = sqrt3 /2 #

We can consider the special triangle:
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Where #cos(x)# = Adjacent / Hypotenues

#cos(30 ) = sqrt3 /2 #

Now we must consider the function of #cosx#:

graph{cosx [-5, 5, -2.5, 2.5]}

This is know as an even function:

#cos(-x) = cos(x) #

so #cos(-30 ) = cos(30) = sqrt3 / 2 #

Answer 2 · 2017-12-11T19:48:41

#cos(-30)^@=sqrt3/2#

#"note that "cos(-30)^@" is in the fourth quadrant"#

#"where "cosx=cos(-x)#

#rArrcos(-30)^@=cos30^@=sqrt3/2#