Question

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

Answer 1

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`