Recursion

We first prove that {b_n} is a convergent sequence.

We have

|b_{n+1}-b_n| = |\cos b_n - \cos b_{n-1}| =2 \left|\sin \left(\frac{b_n-b_{n-1}}{2} \right) \right| \left|\sin \left(\frac{b_n+b_{n-1}}{2} \right) \right|

Now, its obvious that for all n>1, b_n < 1

Hence \left|\sin \left(\frac{b_n+b_{n-1}}{2} \right) \right| < \sin 1

Also \left|\sin \left(\frac{b_n-b_{n-1}}{2} \right) \right| < \left|\left(\frac{b_n-b_{n-1}}{2} \right) \right|

From the above, we have |b_{n+1}-b_n|< |b_n-b_{n-1}|\sin 1 and hence |b_{n+1}-b_n| < |b_1-1| \sin^{n-1}1

That means |b_{n+1}-b_n| may be made as small as we please and hence {b_n} is a Cauchy sequence which is convergent say to l.

Hence we have that l satisfies the equation l = \cos l. (or you can similarly argue that y_n = x_n - \cos x_n converges to 0)

Since f(x) = x - \cos x is monotonic in (0,1), and f \left(\frac{\pi}{6} \right) >0 and f \left(\frac{\pi}{4} \right) <0, there is a root lying in \left(\frac{\pi}{6},\frac{\pi}{4}\right) which fits Option D.

This is in fact the general procedure for finding fixed points of f(x), i.e. solutions for f(x)=x for continuous f

See http://en.wikipedia.org/wiki/Fixed_point_iteration which in fact gives this problem as example #2. Also if you follow the link on Banach fixed point theorem, you will see that the proof here is constructed on similar lines as the one given on that page.

Less rigorous but more intuitive approach.

Starting at 1, follow the black piecewise straight path. Its intersections with the graph of cos x give the subsequent sequence values b₂, b₃, . . .. One can see that this sequence converges towards the point given by the equation x = cos x (the fixed point of the function cos x).