Thread: pre-calc

got it thanks

1. let's convert the angle to degrees pi/8 -> 45/2
now use the formula for double angle -->
cos(2*theta) = 2*[cos(theta/2)]^2 - 1
it's well known that cos(45) = 1/sqrt(2)

thus we get--> cos(45) = 2*[cos(45/2)]^2 - 1

<=> cos(45/2) = sqrt[0.5*(cos(45)+1)] = sqrt[(1+sqrt(2))/(2*sqrt(2))]

2. 2*cos(theta) = 1

<=> theta = arccos(0.5) + 2*pi*k = pi/3 + 2*pi*k
and also (5/3)*pi + 2*pi*k
where k = 0,-1,1,2,-2,....

3. simple enough to me
4. -1-i = sqrt(2)e^(j7*pi/4)

thus (-1-i)^4/3 = [2^(2/3)]e(j*7*pi/3) = use Euler identity...

5. convert to polar form as in the previous section then convert back to Cartesian...