Solution to review question 6.1.4
(i)− 2 p −p^0 p^2 p t3− 3p
23 sin (t − p 2 )
3 p
2p
23 p
−− 2(ii)−p 0 2 p p4− 4p p p 5 p
6 3 2 3 62 p t
35 p
− 6 −
p
− 2
p
− 3
p
− 64 cos ( 2 t − p 6 )
(iii)−p 0 p ttan (t + (^) )
p
2
p
2
p
2
−