Here's the question you clicked on:
agillogly
write (cos pi/4+ i sin pi/4)^3 in trigonometric form
multiply the angle by three
\[\cos(\frac{3\pi}{4})+i\sin(\frac{3\pi}{4})\]
you also have to cube the modulus, but since it is one you just get one, so you can ignore that part now if you want to write this in standard form as \(a+bi\) evaluate the functions and see what you get
\[(\cos(\theta)+i\sin(\theta))^n=\cos(n\theta)+i\sin(n\theta)\] that is what makes this so easy