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STEP III 2007 question 6 solution
From The Student RoomTSR Wiki > Study Help > Subjects and Revision > Mathematics > STEP > STEP III 2007 question 1 solution Firstly,
and similarly for q. Thus we have
and the first result follows. Then, note that if
or
Either way, we have
and thus
from which it follows that
as required. For the next part, we have in the case
Multiplying the first and the third equations, and then dividing by the second, we get
there are obviously two distinct For
In this case, any For larger n, we have exactly the same sitution as in n = 3 when n is odd (you can use the same reasoning, just alternatingly multiply and divide all the equations together) and as in n = 4 when n is even (again the same reasoning, the last equation can always be obtained by alternatingly multiplying and dividing the first ones together. Solution by ukgea. |












and
are perpendicular, 








: (let
represent
in the Argand plane, with the circle centred in the origin and with radius a) 

that satisfy this, as required.
, it's instead
so that the first equation is satisfied, and then a
so that the second one is satisfied, and then a
so that the third is satisfied, and then, and here's the catch, equation 4 follows from the other three: (multiply the first and third equations and divide by the fourth) and so the fourth equation will automatically satified. 




