Find \[ \int_{0}^{\theta}\frac{1}{1-a\cos x}\,\mathrm{d}x\,, \] where \(0 < \theta < \pi\) and \(-1 < a < 1.\) Hence show that \[ \int_{0}^{\frac{1}{2}\pi}\frac{1}{2-a\cos x}\,\mathrm{d}x=\frac{2}{\sqrt{4-a^{2}}}\tan^{-1}\sqrt{\frac{2+a}{2-a}}\,, \] and also that \[ \int_{0}^{\frac{3}{4}\pi}\frac{1}{\sqrt{2}+\cos x}\,\mathrm{d}x=\frac{\pi}{2}\,. \]
Solution: Let \(t = \tan \tfrac{x}{2}\), then \(\cos x = \frac{1-t^2}{1+t^2}, \frac{d t}{d x} =\tfrac12 (1+t^2)\) so the integral is: \begin{align*} \int_0^{\theta} \frac{1}{1-a \cos x} \d x &= \int_{0}^{\tan \frac{\theta}{2}} \frac{1}{1-a \left (\frac{1-t^2}{1+t^2} \right)} \frac{2}{1+t^2} \d t \\ &= \int_0^{\tan \frac{\theta}{2}} \frac{2}{1+t^2-a+at^2} \d t \\ &= \int_0^{\tan \frac{\theta}{2}} \frac{2}{1-a+(1+a) t^2} \d t \\ &= \frac{2}{1+a}\int_0^{\tan \tfrac{\theta}{2}} \frac{1}{\left (\frac{1-a}{1+a} \right)+t^2} \d t \\ &= \frac{2}{1+a} \sqrt{\frac{1+a}{1-a}} \tan^{-1} \left ( \sqrt{\frac{1+a}{1-a}} \tan \frac{\theta}{2} \right) + C \\ &= \frac{2}{\sqrt{1-a^2}}\tan^{-1} \left ( \sqrt{\frac{1+a}{1-a}} \tan \frac{\theta}{2} \right) + C \end{align*} Therefore \begin{align*} \int_{0}^{\frac{1}{2}\pi}\frac{1}{2-a\cos x}\,\mathrm{d}x &= \frac12 \int_0^{\frac12 \pi} \frac{1}{1-\tfrac{a}{2} \cos x} \d x \\ &= \left [\frac12 \frac{2}{\sqrt{1-\frac{a^2}{4}}} \tan^{-1} \left ( \sqrt{\frac{1+\frac{a}{2}}{1-\frac{a}{2}} } \tan\frac{\theta}{2} \right) \right]_0^{\pi/2} \\ &= \frac12 \frac{2}{\sqrt{1-\frac{a^2}{4}}} \tan^{-1} \left ( \sqrt{\frac{1+\frac{a}{2}}{1-\frac{a}{2}} } \tan\frac{\pi}{4} \right) \\ &= \frac{2}{\sqrt{4-a^2}} \tan^{-1} \left ( \sqrt{\frac{2+a}{2-a} } \right) \\ \end{align*} \begin{align*} \int_{0}^{\frac{3}{4}\pi}\frac{1}{\sqrt{2}+\cos x}\,\mathrm{d}x &= \frac{1}{\sqrt{2}} \int_0^{\frac34 \pi} \frac{1}{1 -\left(- \frac{1}{\sqrt{2}} \right)\cos x} \d x \\ &= \frac{1}{\sqrt{2}} \left [ \frac{2}{\sqrt{1-\tfrac12}} \tan^{-1} \left ( \sqrt{\frac{1-\frac{1}{\sqrt{2}}}{1+\frac{1}{\sqrt{2}}} } \tan\frac{\theta}{2} \right) \right]_0^{3\pi/4} \\ &= \frac{1}{\sqrt{2}} \frac{2}{\sqrt{1/2}} \tan^{-1} \left ( \sqrt{\frac{\sqrt{2}-1}{\sqrt{2}+1} } \tan\frac{3\pi}{8} \right) \\ &= 2 \tan^{-1} \left ( \sqrt{\frac{(\sqrt{2}-1)^2}{2-1} } \tan\frac{3\pi}{8} \right)\\ &= 2 \tan^{-1} \left ( (\sqrt{2}-1) \tan\frac{3\pi}{8} \right) \end{align*} If \(t = \tan \tfrac{3\pi}{8}\), then \(-1 = \tan \tfrac{3\pi}{4} = \frac{2t}{1-t^2} \Rightarrow t^2-2t-1 = 0 \Rightarrow t = 1\pm \sqrt{2}\), since \( t > 0\), we must have \(t = 1 + \sqrt{2}\), so \begin{align*} \int_{0}^{\frac{3}{4}\pi}\frac{1}{\sqrt{2}+\cos x}\,\mathrm{d}x &= 2 \tan^{-1} \left ((\sqrt{2}-1)(\sqrt{2}+1) \right) \\ &= 2 \tan^{-1} 1 \\ &= 2 \frac{\pi}{4} \\ &= \frac{\pi}{2} \end{align*}
Find the integers \(k\) satisfying the inequality \(k\leqslant2(k-2).\) Given that \(N\) is a strictly positive integer consider the problem of finding strictly positive integers whose sum is \(N\) and whose product is as large as possible. Call this largest possible product \(P(N).\) Show that \(P(5)=2\times3,P(6)=3^{2},P(7)=2^{2}\times3,P(8)=2\times3^{2}\) and \(P(9)=3^{3}.\) Find \(P(1000)\) explaining your reasoning carefully.
Show, using de Moivre's theorem, or otherwise, that \[ \tan7\theta=\frac{t(t^{6}-21t^{4}+35t^{2}-7)}{7t^{6}-35t^{4}+21t^{2}-1}\,, \] where \(t=\tan\theta.\)
Solution:
Solution:
A transformation \(T\) of the real numbers is defined by \[ y=T(x)=\frac{ax-b}{cx-d}\,, \] where \(a,b,c\), \(d\) are real numbers such that \(ad\neq bc\). Find all numbers \(x\) such that \(T(x)=x.\) Show that the inverse operation, \(x=T^{-1}(y)\) expressing \(x\) in terms of \(y\) is of the same form as \(T\) and find corresponding numbers \(a',b',c'\),\(d'\). Let \(S_{r}\) denote the set of all real numbers excluding \(r\). Show that, if \(c\neq0,\) there is a value of \(r\) such that \(T\) is defined for all \(x\in S_{r}\) and find the image \(T(S_{r}).\) What is the corresponding result if \(c=0\)? If \(T_{1},\) given by numbers \(a_{1},b_{1},c_{1},d_{1},\) and \(T_{2},\) given by numbers \(a_{2},b_{2},c_{2},d_{2}\) are two such transformations, show that their composition \(T_{3},\) defined by \(T_{3}(x)=T_{2}(T_{1}(x)),\) is of the same form. Find necessary and sufficient conditions on the numbers \(a,b,c,d\) for \(T^{2}\), the composition of \(T\) with itself, to be the identity. Hence, or otherwise, find transformations \(T_{1},T_{2}\) and their composition \(T_{3}\) such that \(T_{1}^{2}\) and \(T_{2}^{2}\) are each the identity but \(T_{3}^{2}\) is not.
A particle of mass \(m\) is at rest on top of a smooth fixed sphere of radius \(a\). Show that, if the particle is given a small displacement, it reaches the horizontal plane through the centre of the sphere at a distance % at least $$a(5\sqrt5+4\sqrt23)/27$$ from the centre of the sphere. [Air resistance should be neglected.]
Two rough solid circular cylinders, of equal radius and length and of uniform density, lie side by side on a rough plane inclined at an angle \(\alpha\) to the horizontal, where \(0<\alpha<\pi/2\). Their axes are horizontal and they touch along their entire length. The weight of the upper cylinder is \(W_1\) and the coefficient of friction between it and the plane is \(\mu_1\). The corresponding quantities for the lower cylinder are \(W_2\) and \(\mu_2\) respectively and the coefficient of friction between the two cylinders is \(\mu\). Show that for equilibrium to be possible:
Solution:
A smooth circular wire of radius \(a\) is held fixed in a vertical plane with light elastic strings of natural length \(a\) and modulus \(\lambda\) attached to the upper and lower extremities, \(A\) and \(C\) respectively, of the vertical diameter. The other ends of the two strings are attached to a small ring \(B\) which is free to slide on the wire. Show that, while both strings remain taut, the equation for the motion of the ring is $$2ma \ddot\theta=\lambda(\cos\theta-\sin\theta)-mg\sin\theta,$$ where \(\theta\) is the angle \( \angle{CAB}\). Initially the system is at rest in equilibrium with \(\sin\theta=\frac{3}{5}\). Deduce that \(5\lambda=24mg\). The ring is now displaced slightly. Show that, in the ensuing motion, it will oscillate with period approximately $$10\pi\sqrt{a\over91g}\,.$$
It has been observed that Professor Ecks proves three types of theorems: 1, those that are correct and new; 2, those that are correct, but already known; 3, those that are false. It has also been observed that, if a certain of her theorems is of type \(i\), then her next theorem is of type \(j\) with probability \(p\low_{ij},\) where \(p\low_{ij}\) is the entry in the \(i\)th row and \(j\)th column of the following array: \[ \begin{pmatrix}0.3 & 0.3 & 0.4\\ 0.2 & 0.4 & 0.4\\ 0.1 & 0.3 & 0.6 \end{pmatrix}\,. \] Let \(a_{i},\) \(i=1,2,3\), be the probability that a given theorem is of type \(i\), and let \(b_{j}\) be the consequent probability that the next theorem is of type \(j\).