Find  cos 72o  using complex number

 

 

 

Some trigonometric function of special angles can be evaluated using the concept of complex number.  Here is an example.

 

Put                z = cos 72o + i sin 72o .

Then               z5 = cos 360o + i sin 360o = 1

Therefore        z5 – 1 = 0

                       (z – 1)(z4 + z3 + z2 + z + 1) = 0

Since  z ¹ 1,   z4 + z3 + z2 + z + 1 = 0

                      

                      

                      

                        

 

Solving this quadratic equation, and rejecting the negative root, we have

                      

Note that: