STEP III
Q5.
(Here, q denotes theta, c denotes cos(q) and s denotes sin(q).)
(cos(q) + isin(q))^7 = cos(7q) + isin(7q)
Expand the LHS using the binomial theorem, then equate real and imaginary parts to get
cos(7q) = c^7 - 21 c^5 s^2 + 35 c^3 s^4 - 7 c s^6
sin(7q) = 7 c^6 s - 35 c^4 s^3 + 21 c^2 s^5 - s^7
Thus
tan(7q) = (7 c^6 s - 35 c^4 s^3 + 21 c^2 s^5 - s^7)/(
c^7 - 21 c^5 s^2 + 35 c^3 s^4 - 7 c s^6)
= (7t - 35t^3 + 21t^5 - t^7)/(1 - 21t^2 + 35t^4 - 7t^6)
= t(t^6 - 21t^4 + 35t^2 - 7)/(7t^6 - 35t^4 + 21t^2 - 1), as required.
(i)
If q=pi/7, 2pi/7, 3pi/7, then tan(7q)=0 while tan(q) != 0. So,
0 = tan^6(q) - 21tan^4(q) + 35tan^2(q) - 7
That is, tan^2(pi/7), tan^2(2pi/7) and tan^2(3pi/7) are roots of the equation
x^3 - 21x^2 + 35x - 7 = 0
The product of the roots of this equation is 7. That is,
tan^2(pi/7) tan^2(2pi/7) tan^2(3pi/7) = 7
(ii)
This time we want to find a cubic whose roots are tan^2(pi/14), tan^2(3pi/14) and tan^2(5pi/14). Notice that if q=pi/14, 3pi/14, 5pi/14, then tan(7q) is undefined while tan(q) is. So,
7tan^6(q) - 35tan^4(q) + 21tan^2(q) - 1 = 0
will do the trick.
Thus, the 3 values of tan are roots of the equation
7x^3 - 35x^2 + 21x^2 - 1 = 0
The sum of the roots of this equation is 35/7 = 5. That is,
tan^2(pi/14) + tan^2(3pi/14) + tan^2(5pi/14) = 5
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Q6.
(Group theory is no longer on the A-level syllabus, so this question is more or less irrelevant now. I'll do it anyway.)
(i)
Let A be the 2x2 matrix whose entries are all 1. Note that A^2 = 2A.
So S is simply { xA : x is a nonzero real }. Let's show that it's closed under matrix multiplication. Let xA and yA be in S. Then:
xA * yA = xy A^2 = 2xy A, which is in S.
To show that S forms a group under matrix multiplication, first we note that matrix multiplication is associative. We take the identity element to be (1/2)A. Indeed, if xA is in S then 1/2 A * xA = xA * 1/2 A = x/2 A^2 = x A. And, finally, if yA is in S, then because y is nonzero, we have that 1/(2y) A is its inverse. Thus all the axioms are satisfied.
(ii)
Suppose A in G is singular. Since G is a group, A has a group inverse B. That is, AB = E; whence,
det(E) = det(AB) = det(A)det(B) = 0
So det(E) = 0. Thus, if C is any matrix in G, then det(C) = det(CE) = det(C)det(E) = 0, and therefore C is singular.
For an example, look back at S in part (i). This time take the set of 3x3 matrices of that form - call it T. Here, if we let B be the 3x3 matrix whose entries are all 1, then B^2 = 3B. So showing that T is a group is analagous to show that S is. The identity in this case is B/3, which is a singular matrix.
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Q8.
(I don't think this question is relevant either.)
If T(x) = x, then
ax - b = x(cx - d)
=> ax - b = cx^2 - dx
=> cx^2 - (a+d)x + b = 0
=> x = [(a+d) +/- sqrt((a+d)^2 - 4bc)]/2c
If T(x) = y, then
ax - b = y(cx - d)
=> ax - b = cxy - dy
=> ax - cxy = b - dy
=> x(a - cy) = b - dy
=> x = (b - dy)/(a - cy)
So T-1(x) = (dx - b)/(cx - a). Thus, a'=d, b'=b, c'=c and d'=a.
If c!=0, then T(x) always gives us a real number unless cx-d=0 => x=d/c. So we take r=d/c. To find the image of S_r, let x be in S_r and let y = T(x). Then,
x = (dy - b)/(cy - a)
So y != a/c. Thus, T(S_r) = { x in R : x != a/c }.
On the other hand, if c=0, then T(x) = (-a/d)x + b/d. So if d=0 then we have a problem; otherwise, T is defined for all real numbers x. So T(S_r) is all of R minus the point (-a/d)r + b/d.
Moving on.
T_1(T_2(x)) = [(a_1 a_2 - b_1 c_2)x - (a_1 b_2 - b_1 d_2)]/[(c_1 a_2 - d_1 c_2)x - (c_1 b_2 - d_1 d_2)], so it's of the same form.
Thus, T^2(x) is the identity iff the following conditions are satisfied:
a^2 - bc = 1
ab - bd = 0
ac - dc = 0
bc - d^2 = -1
Finally, for the last bit, if we have for instance (a_3)^2 - b_3 c_3 = 0, then T_3 is not the identity. Let's re-express this:
(a_3)^2 = b_3 c_3
=> (a_1 a_2 - b_1 c_2)^2 = (a_1 b_2 - d_1 d_2) (c_1 a_2 - d_1 c_2)
Now let's take T_1(x) = 1-x. That is,
a_1 = b_1 = -1, c_1 = d_1 = 0
Then T_2 must satisfy the extra constraint
(c_2 - a_2)^2 = b_2
Let's take a_2 = 0. Then (c_2)^2 = b_2, and b_2 c_2 = -1. So b_2 = 1 and c_2 = -1; also, b_2 c_2 - (d_2)^2 = -1 implies d_2 = 0.
Thus T_2(x) = -1/x, and T_3(x) = 1 + 1/x.
Let's verify:
T_1(x)^2 = 1 - (1 - x) = x
T_2(x)^2 = -1/(-1/x) = x
T_3(x)^2 = 1 + 1/(1 + 1/x) = 1 + x/(1 + x) != x
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Is it just me or is III/1 unbearably long?