1. cos(3x) = cos(2x+x) = cos(2x)cos(x)-sin(2x)sin(x) = (2cos˛(x)-1)cos(x)-2sin(x)cos(x)sin(x) = 2cosł(x)-cos(x)-2sin˛(x)cos(x) = 2cosł(x)-cos(x)-2[1-cos˛(x)]cos(x) = 2cosł(x)-cos(x)-2cos(x)+2cosł(x) 4cosł(x)-3cos(x) So cos(3x)=4cosł(x)-3cos(x) and cos(2x)=2cos˛(x)-1. So cos(6x)=2[cos(3x)]˛-1 or 2[4cosł(x)-3cos(x)]˛-1. We can expand it to become 2[16cos6(x)-24cos4(x)+9cos˛(x)]-1 Final result: cos(6x) = 32cos6(x) - 48cos4(x) + 18cos2(x) - 1 ------------------------ 2.---------------------- This one requires some tricks of subtracting and adding the same quantity to create a use for some double-angle identities. We work with the left side: Subtract 1 and add 1 to create a use for the identity Using that identity, the first two terms become cos(6x) Subtract and add cos(7x)cos(x) to create a use for the identity Factor a "-" out of 3rd and 4th terms: Swap the terms in the parentheses to recognize the identity: Use the identity 7x-x = 6x Cancel the two opposite terms Tricky, huh? Edwin