We recall the 30-60 triangle from before:

30-60 degree triangle

Using

cos(α + β) = cos α cos β − sin α sin β

and our 30-60 triangle, we start with the left hand side (LHS) and obtain:

`"LHS"=cos(30^"o"+x)`

`=cos\ 30^"o"\ cos\ x - sin\ 30^"o"\ sin\ x`

`=sqrt3/2 cos\ x-1/2sin\ x`

`=(sqrt3\ cos\ x-sin\ x)/2`

`="RHS"`

Since the LHS = RHS, we have proved the identity.