Show that #cot (-390°)-4 cos 480# = #2-sqrt3#.?

1 Answer
Mar 27, 2018

See the method below.

Explanation:

#cot(-390˚)-4cos(480˚)=2-sqrt3#

Recall that #cot(x) = 1/tanx = cosx/sinx#
Since all trigonometric functions repeat after 360˚ we can add or subtract 360˚ to the angles without changing anything. Doing this we get

#cot(-360˚ -30˚)-4cos(360˚ + 120˚)=2-sqrt3=#
#cot(-30˚)-4cos(120˚)=2-sqrt3#

We can write #cotx# as #cosx/sinx#

#cos(-30˚)/sin(-30˚)-4cos(120˚)=2-sqrt3#

This can be further simplified as #cos(-x)= cos(x)# and #sin(-x)=-sin(x)#.

#-cos(30˚)/sin(30˚)-4cos(120˚)=2-sqrt3#

Consulting the unit circle we find the following:

#cos(30˚) = sqrt3/2 #
#sin(30˚) = 1/2#
#cos(120˚)=-1/2#

We can now substitute these values in.

#-(sqrt3/2)/{1/2}-4(-1/2)=2-sqrt3#

Simplifying we get

#-(sqrt3)/{1}+4/2=2-sqrt3=#

#-sqrt3+2=2-sqrt3#

Rearranging we get #2-sqrt3=2-sqrt3#