rcos2theta2Solution First notice that we can use the doublea

Solution

First, notice that we can use the double-angle formula for cosine: cos(2?) = 2cos^2? - 1, to re-write cos^2(?/2) in a nicer form. Solving the above equation for cos^2? gives: cos^2? = [1 + cos(2?)]/2. So, replacing ? with ?/2 gives. cos^2(?/2) = (1 + cos?)/2. Recall that the length of a polar curve of the form r = f(?) on [a, b] is given by: L = ? v[r^2 + (dr/d?)^2] d? (from ?=a to b). You haven\'t given us the intervals, so I\'ll leave the calculations to you. Just plug in r, dr/d?, and the bounds and integrate.