This question is concerned with an l2S spin system.

(b) By first applying the identity cos2 (A) = ½ j[l + cos(2A)], and then the identity cos (A)cos(B) = 1/2 [cos (A + B) + cos (A - B)], show that the product of trigonometric terms in front of the operator Sx can be written as the following sum of trigono¬metric terms; you may also need to use cos(-A) = cos (A)
1/4 [2cos(Ωst) + cos(Ωst + 2πJt) + cos(Ωst - 2πJt).

(c) Assuming that only the x-magnetization is detected and the resulting signal is subject to a cosine Fourier transform, ex¬plain why your result predicts that the spectrum will consist of a 1:2:1 triplet, centred at Qs- What is the spacing of the outer two lines?
(d) Repeat the above calculation for the case where the initial state is 2Izx and show that the resulting term in is
- sin (Ωst) cos (πJt) sin (πJt) x.
By using the identities sin (A) cos (A) = j sin (2A) and then sin (A) sin (B) = 1/2 [cos(A - B) - cos (A + B)], show that the product of trigonometric terms can be written as
1/4 [cos (Ωst + 2πJt) - cos (Ωst - 2πJt)].
Hence explain that the resulting spectrum is a -1:0:+1 triplet.
(e) Repeat the above calculation for the case where the ini¬tial state is 41z2Zx and show that the resulting spectrum is a +l:-2:+l triplet. You may need to use the identity sin2 (A) = 1/2 [l-cos(2A)].

cos (2st) cos (7Jt)-