Trigonometry

Let  the sides be  a, b and c

So we have that

a^2 =  b^2 + c^2 - 2bc *cos (theta)     rearrange as

2bc * cos(theta)  =  b^2 + c^2 - a^2      divide both sides by   2bc

cos (theta)  =  [  b^2 + c^2 - a^2 ] /  [2bc ]

Take the cosine inverse

cos^-1 (  [  b^2 + c^2 - a^2 ] /  [2bc ] )  =  theta

Remember that "theta"  is the angle opposite  side "a"

Notice from this point 

a^2 =  b^2 + c^2 - 2bc *cos (theta)  

I subtracted a^2  from both sides and added 2bc *cos (theta)   to both sides giving us

2bc * cos(theta)  =  b^2 + c^2 - a^2 

Your rearrangement is correct if  side "c"  is opposite "theta"

In my answer, I arbitrarily made  "a"  the opposite side to "theta"