Express cos 5x in terms of sine multiples of x

cos x = [ e^(ix) - e^(-ix) ] / 2i along with [ A - B ]^5 = 

( A^5 - B^5) - 5 AB ( A^3 - B^3) + 10 A²B² (A - B )...A = e^(ix) , B = e^(-ix) 

[ cos x ] ^5 = (1/32i) { [ e^(5ix) - e^(-5x) ] - 5 [ ( e^(3ix) - e^(-3ix) ] + 10 [ e^(ix) - e^(-ix) ] } 

= ( 1 / 16 ) { cos 5x - 5 cos 3x + 10 cos x }much faster