There has been a discussion of some of the harder stuff that can turn up at the ends of papers, so I thought I'd post some stuff from later questions of some 4u half-yearlies I have. I've skipped the motion, as people probably haven't done it yet, but please advise if I'm wrong...
1. (For this question, I'm going to use a in place of alpha, for the sake of clarity.)
(a) Provided sin(a / 2) <> 0, show that cos(a / 2) = sin a / 2sin(a / 2)
Similarly, show that if sin(a / 2) and sin(a / 4) are both <> 0, cos(a / 2) * cos(a / 4) = sin a / 4sin(a / 4)
(b) Prove by mathematical induction that if n is a positive integer, and sin(a / 2<sup>n</sup>) <> 0, then
cos(a / 2) * cos(a / 4) * ... * cos(a / 2<sup>n</sup>) = sin a / 2<sup>n</sup>sin(a / 2<sup>n</sup>)
(c) Hence, deduce that a = sin a / {[cos(a / 2)] * [cos(a / 4)] * [cos(a / 8)] * ... } and that
pi = 1 / {(1/2) * sqrt(1/2) * sqrt[(1/2) + (1/2)sqrt(1/2)] * sqrt{(1/2) + (1/2)sqrt[(1/2) + (1/2)sqrt(1/2)]} * ...}
2. (a) Let @ = tan<sup>-1</sup>x + tan<sup>-1</sup>y. Show that tan@ = (x + y) / (1 - xy)
(b) If tan<sup>-1</sup>x + tan<sup>-1</sup>y + tan<sup>-1</sup>z = pi / 2, show that xy + yz + zx = 1
(c) Let W<sub>n</sub> = tan<sup>-1</sup>x<sub>1</sub> + tan<sup>-1</sup>x<sub>2</sub> + ... + tan<sup>-1</sup>x<sub>n</sub>, where n is a positive integer.
Show, by mathematical induction or otherwise, that tan W<sub>n</sub> = - Im(w<sub>n</sub>) / Re(w<sub>n</sub>)
where w<sub>n</sub> = (1 - ix<sub>1</sub>)(1 - ix<sub>2</sub>)...(1 - ix<sub>n</sub>)
3. The quartic polynomial f(x) = x<sup>4</sup> + bx<sup>3</sup> + cx<sup>2</sup> + dx + e has two double zeroes at alpha and beta, where the coefficients b, c, d, and e are real, but the zeroes alpha and beta may be complex.
(a) Express b, c, d and e in terms of the double zeroes alpha and beta, and hence show that:
(i) d<sup>2</sup> = b<sup>2</sup>e
(ii) b<sup>3</sup> + 8d = 4bc
(b) Now, suppose that b = 2 and e = 1, and that the double zeroes alpha and beta are both non-real complex numbers.
(i) Show that alpha and beta are cube roots of 1.
(ii) Write f(x) as the product of polynomials irreducible over the real numbers.
1. (For this question, I'm going to use a in place of alpha, for the sake of clarity.)
(a) Provided sin(a / 2) <> 0, show that cos(a / 2) = sin a / 2sin(a / 2)
Similarly, show that if sin(a / 2) and sin(a / 4) are both <> 0, cos(a / 2) * cos(a / 4) = sin a / 4sin(a / 4)
(b) Prove by mathematical induction that if n is a positive integer, and sin(a / 2<sup>n</sup>) <> 0, then
cos(a / 2) * cos(a / 4) * ... * cos(a / 2<sup>n</sup>) = sin a / 2<sup>n</sup>sin(a / 2<sup>n</sup>)
(c) Hence, deduce that a = sin a / {[cos(a / 2)] * [cos(a / 4)] * [cos(a / 8)] * ... } and that
pi = 1 / {(1/2) * sqrt(1/2) * sqrt[(1/2) + (1/2)sqrt(1/2)] * sqrt{(1/2) + (1/2)sqrt[(1/2) + (1/2)sqrt(1/2)]} * ...}
2. (a) Let @ = tan<sup>-1</sup>x + tan<sup>-1</sup>y. Show that tan@ = (x + y) / (1 - xy)
(b) If tan<sup>-1</sup>x + tan<sup>-1</sup>y + tan<sup>-1</sup>z = pi / 2, show that xy + yz + zx = 1
(c) Let W<sub>n</sub> = tan<sup>-1</sup>x<sub>1</sub> + tan<sup>-1</sup>x<sub>2</sub> + ... + tan<sup>-1</sup>x<sub>n</sub>, where n is a positive integer.
Show, by mathematical induction or otherwise, that tan W<sub>n</sub> = - Im(w<sub>n</sub>) / Re(w<sub>n</sub>)
where w<sub>n</sub> = (1 - ix<sub>1</sub>)(1 - ix<sub>2</sub>)...(1 - ix<sub>n</sub>)
3. The quartic polynomial f(x) = x<sup>4</sup> + bx<sup>3</sup> + cx<sup>2</sup> + dx + e has two double zeroes at alpha and beta, where the coefficients b, c, d, and e are real, but the zeroes alpha and beta may be complex.
(a) Express b, c, d and e in terms of the double zeroes alpha and beta, and hence show that:
(i) d<sup>2</sup> = b<sup>2</sup>e
(ii) b<sup>3</sup> + 8d = 4bc
(b) Now, suppose that b = 2 and e = 1, and that the double zeroes alpha and beta are both non-real complex numbers.
(i) Show that alpha and beta are cube roots of 1.
(ii) Write f(x) as the product of polynomials irreducible over the real numbers.
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