137 problems found
Prove the identity \[ 4\sin\theta \sin(\tfrac13\pi-\theta) \sin (\tfrac13\pi+\theta)= \sin 3\theta\, . \tag{\(*\)}\]
Solution: \begin{align*} && LHS &= 4\sin\theta \sin(\tfrac13\pi-\theta) \sin (\tfrac13\pi+\theta) \\ &&&= 4 \sin \theta \left (\tfrac{\sqrt{3}}{2}\cos \theta - \tfrac12 \sin \theta \right)\left (\tfrac{\sqrt{3}}{2}\cos \theta + \tfrac12 \sin \theta \right) \\ &&&= 4 \sin \theta \left (\tfrac{3}{4}\cos^2 \theta - \tfrac14 \sin^2 \theta \right) \\ &&&= 3\sin \theta - 4\sin^3 \theta \\ &&&= \cos 3 \theta = RHS \end{align*}
Write down the cubes of the integers \(1, 2, \ldots , 10\). The positive integers \(x\), \(y\) and \(z\), where \(x < y\), satisfy \[ x^3+y^3 = kz^3\,, \tag{\(*\)} \] where \(k\) is a given positive integer.
Solution: \begin{array}{c|c} n & n^3 \\ \hline 1 & 1 \\ 2 & 8 \\ 3 & 27 \\ 4 & 64 \\ 5 & 125 \\ 6 & 216 \\ 7 & 343 \\ 8 & 512 \\ 9 & 729 \\ 10 & 1000 \\ \end{array}
Show that \[ \sin(x+y) -\sin(x-y) = 2 \cos x \, \sin y \] and deduce that \[ \sin A - \sin B = 2 \cos \tfrac12 (A+B) \, \sin\tfrac12 (A-B) \,. \] Show also that \[ \cos A - \cos B = -2 \sin \tfrac12(A+B) \, \sin\tfrac12(A-B)\,. \] The points \(P\), \(Q\), \(R\) and \(S\) have coordinates \(\left(a\cos p,b\sin p\right)\), \(\left(a\cos q,b\sin q\right)\), \(\left(a\cos r,b\sin r\right)\) and \(\left(a\cos s,b\sin s\right)\) respectively, where \(0\le p < q < r < s <2\pi\), and \(a\) and \(b\) are positive. Given that neither of the lines \(PQ\) and \(SR\) is vertical, show that these lines are parallel if and only if \[ r+s-p-q = 2\pi\,. \]
Solution: \begin{align*} && \sin(x+y) - \sin(x-y) &= \sin x \cos y + \cos x \sin y - (\sin x \cos y - \cos x \sin y )\\ &&&= 2 \cos x \sin y \\ \\ && A &= x+y \\ && B &= x - y \\ \Rightarrow && x = \frac12(A+B) &\quad y = \frac12(A-B) \\ \Rightarrow && \sin A - \sin B &= 2 \cos \tfrac12(A+B) \sin \tfrac12(A-B) \\ \\ && \cos (x+y) - \cos (x-y) &= \cos x \cos y - \sin x \sin y -(\cos x \cos y + \sin x \sin y ) \\ &&&= -2 \sin x \sin y \\ \Rightarrow && \cos A - \cos B &= - 2 \sin \tfrac12 (A+B) \sin \tfrac12 (A-B) \end{align*} \begin{align*} && \text{Gradient of }PQ &= \frac{b \sin q - b \sin p}{a \cos q - a \cos p } \\ && \text{Gradient of }SR &= \frac{b \sin s - b \sin r}{a \cos s - a \cos r} \\ PQ \parallel SR \Rightarrow && \frac{b \sin q - b \sin p}{a \cos q - a \cos p } &= \frac{b \sin s - b \sin r}{a \cos s - a \cos r} \\ \Rightarrow && (\sin q - \sin p)(\cos s - \cos r) &= (\sin s - \sin r)(\cos q - \cos r) \\ \Rightarrow && -4 \cos \tfrac12(p+q) \sin\tfrac12(q-p) \sin \tfrac12(s+r) \sin \tfrac12(s-r) &= -4 \cos \tfrac12(s+r) \sin \tfrac12(s-r) \sin \tfrac12 (p+q) \sin\tfrac12 (q-p) \\ \Rightarrow && 0 &= \sin \tfrac12(s-r)\sin\tfrac12(p-q) \left ( \cos \tfrac12(p+q)\sin \tfrac12(s+r) - \sin \tfrac12 (p+q)\cos \tfrac12(s+r) \right) \\ &&&= \sin \tfrac12(s-r)\sin\tfrac12(p-q) \sin \left ( \frac12 (s+r -(p+q))\right) \end{align*} Since \(s \neq r\) and \(p \neq q\) (neither line vertical) we must have \(\frac12 (s+r -(p+q)) = n \pi \Rightarrow s+r - p - q = 0, 2\pi, 4\pi, \cdots\) but given the range constraints it must be \(2 \pi\)
Solution:
Prove that \[ \cos 3x = 4 \cos^3 x - 3 \cos x \,. \] Find and prove a similar result for \(\sin 3x\) in terms of \(\sin x\).
Solution: \begin{align*} \cos 3x &\equiv \cos (2x + x) \\ &\equiv \cos 2x \cos x - \sin 2x \sin x \\ &\equiv (2\cos^2 x - 1) \cos x - 2 \sin x \cos x \sin x \\ &\equiv 2 \cos^3 x - \cos x - 2\cos x (\sin^2 x) \\ &\equiv 2 \cos^3 x - \cos x - 2\cos x (1- \cos^2 x) \\ &\equiv 4\cos^3 x - 3\cos x \end{align*} Similarly, \begin{align*} \sin 3x &\equiv \sin (2x + x) \\ &\equiv \sin 2x \cos x + \cos 2x \sin x \\ &\equiv 2 \sin x \cos x \cos x + (1-2\sin^2 x) \sin x \\ &\equiv 2 \sin x (1-\sin^2 x) + \sin x - 2 \sin^3 x \\ &\equiv 3 \sin x -4 \sin ^3 x \end{align*}
Let \(x_{\low1}\), \(x_{\low2}\), \ldots, \(x_n\) and \(x_{\vphantom {\dot A} n+1}\) be any fixed real numbers. The numbers \(A\) and \(B\) are defined by \[ A = \frac 1 n \sum_{k=1}^n x_{ \low k} \,, \ \ \ B= \frac 1 n \sum_{k=1}^n (x_{\low k}-A)^2 \,, \ \ \ \] and the numbers \(C\) and \(D\) are defined by \[ C = \frac 1 {n+1} \sum\limits_{k=1}^{n+1} x_{\low k} \,, \ \ \ D = \frac1{n+1} \sum_{k=1}^{n+1} (x_{\low k}-C)^2 \,. \]
Solution:
The infinite series \(S\) is given by \[ S = 1 + (1 + d)r + (1 + 2d)r^2 + \cdots + (1+nd)r^n +\cdots\; ,\] for \(\vert r \vert <1\,\). By considering \(S - rS\), or otherwise, prove that \[ S = \frac 1{1-r} + \frac {rd}{(1-r)^2} \,.\] Arthur and Boadicea shoot arrows at a target. The probability that an arrow shot by Arthur hits the target is \(a\); the probability that an arrow shot by Boadicea hits the target is \(b\). Each shot is independent of all others. Prove that the expected number of shots it takes Arthur to hit the target is \(1/a\). Arthur and Boadicea now have a contest. They take alternate shots, with Arthur going first. The winner is the one who hits the target first. The probability that Arthur wins the contest is \(\alpha\) and the probability that Boadicea wins is \(\beta\). Show that \[ \alpha = \frac a {1-a'b'}\,, \] where \(a' = 1-a\) and \(b'=1-b\), and find \(\beta\). Show that the expected number of shots in the contest is \(\displaystyle \frac \alpha a + \frac \beta b\,.\)
Solution: Notice that \begin{align*} && S - rS &= 1 + dr + dr^2 + \cdots \\ &&&= 1 + dr(1 + r+r^2+ \cdots) \\ &&&= 1 + \frac{rd}{1-r} \\ \Rightarrow && S &= \frac{1}{1-r} + \frac{rd}{(1-r)^2} \end{align*} The number of shots Arthur takes is \(\textrm{Geo}(a)\), so it's expectation is \(1/a\). The probability Arthur wins is: \begin{align*} \alpha &= a + a'b'a + (a'b')^2a + \cdots \\ &= a(1+a'b' + \cdots) \\ &= \frac{a}{1-a'b'} \\ \\ \beta &= a'b + a'b'a'b + \cdots \\ &= a'b(1+b'a' + (b'a')^2 + \cdots ) \\ &= \frac{a'b}{1-a'b'} \end{align*} The expected number of shots in the contest is: \begin{align*} E &= a + 2a'b + 3a'b'a + 4a'b'a'b + \cdots \\ &= a(1 + 3a'b' + 5(a'b')^2 + \cdots) + 2a'b(1 + 2(a'b') + 3(a'b')^2 + \cdots) \\ &= a \left ( \frac{1}{1-a'b'} + \frac{2a'b'}{(1-a'b')^2} \right) + 2a'b \left ( \frac{1}{1-a'b'} + \frac{a'b'}{(1-a'b')^2}\right) \\ &= \frac{a}{1-a'b'} \left (1 + \frac{2a'b'}{(1-a'b')} \right) + 2\frac{a'b}{1-a'b'} \left ( 1 + \frac{a'b'}{(1-a'b')}\right) \\ &= \alpha \frac{1+a'b'}{1-a'b'} + \beta \frac{2}{1-a'b'} \\ &= \alpha \frac{1+1-a-b+ab}{1-a'b'} + \beta \frac{2}{1-a'b'} \\ \end{align*}
The points \(S\), \(T\), \(U\) and \(V\) have coordinates \((s,ms)\), \((t,mt)\), \((u,nu)\) and \((v,nv)\), respectively. The lines \(SV\) and \(UT\) meet the line \(y=0\) at the points with coordinates \((p,0)\) and \((q,0)\), respectively. Show that \[ p = \frac{(m-n)sv}{ms-nv}\,, \] and write down a similar expression for \(q\). Given that \(S\) and \(T\) lie on the circle \(x^2 + (y-c)^2 = r^2\), find a quadratic equation satisfied by \(s\) and by \(t\), and hence determine \(st\) and \(s+t\) in terms of \(m\), \(c\) and \(r\). Given that \(S\), \(T\), \(U\) and \(V\) lie on the above circle, show that \(p+q=0\).
Prove that, if \(c\ge a\) and \(d\ge b\), then \[ ab+cd\ge bc+ad\,. \tag{\(*\)} \]
Solution: \begin{align*} && \underbrace{(c-a)}_{\geq 0}\underbrace{(d-b)}_{\geq 0} & \geq 0 \\ \Leftrightarrow && cd -bc -ad + ab &\geq 0 \\ \Leftrightarrow && ab +cd &\geq bc+ad \\ \end{align*}
The polynomial \(\p(x)\) is given by \[ \ds \p(x)= x^n +\sum\limits_{r=0}^{n-1}a_rx^r\,, \] where \(a_0\), \(a_1\), \(\ldots\) , \(a_{n-1}\) are fixed real numbers and \(n\ge1\). Let \(M\) be the greatest value of \(\big\vert \p(x) \big\vert\) for $\vert x \vert\le 1\(. Then Chebyshev's theorem states that \)M\ge 2^{1-n}$.
Solution:
Solution:
Show that \(x^3-3xbc + b^3 + c^3\) can be written in the form \(\left( x+ b+ c \right) {\rm Q}( x)\), where \({\rm Q}( x )\) is a quadratic expression. Show that \(2{\rm Q }( x )\) can be written as the sum of three expressions, each of which is a perfect square. It is given that the equations \(ay^2 + by + c =0\) and \(by^2 + cy + a = 0\) have a common root \(k\). The coefficients \(a\), \(b\) and \(c\) are real, \(a\) and \(b\) are both non-zero, and \(ac \neq b^2\). Show that \[ \left( ac - b^2 \right) k = bc - a^2 \] and determine a similar expression involving \(k^2\). Hence show that \[ \left( ac - b^2 \right) \left(ab-c^2 \right) = \left( bc - a^2 \right)^2 \] and that \( a^3 -3abc + b^3 +c^3 = 0\,\). Deduce that either \(k=1\) or the two equations are identical.
Solution: \begin{align*} && x^3 - 3xbc+b^3 + c^3 &= (x+b+c)(x^2-x(b+c)+b^2+c^2-bc) \\ &&&= \tfrac12(x+b+c)((x-b)^2+(x-c)^2+(b-c)^2) \\ \end{align*} We must have: \begin{align*} && 0 &= ak^2 + bk+c \tag{1}\\ &&0 &= bk^2+ck+a \tag{2}\\ b*(1)&& 0 &= abk^2 + b^2k+cb \\ a*(2)&& 0 &= abk^2 + ack + a^2 \\ \Rightarrow && 0 &= k(ac-b^2)+a^2-bc \\ \Rightarrow && (ac-b^2)k &= bc-a^2 \\ \\ c*(1) && 0 &= ack^2+bck+c^2 \\ b*(2) && 0 &= b^2k^2+bck+ab \\ \Rightarrow && 0 &= (ac-b^2)k^2 +c^2-ab \\ \Rightarrow && (ac-b^2)k^2 &= ab-c^2 \\ \\ \Rightarrow && \frac{ab-c^2}{ac-b^2} &= k^2 = \left (\frac{bc-a^2}{ac-b^2} \right)^2 \\ \Rightarrow && (ab-c^2)(ac-b^2) &= (bc-a^2)^2 \\ \Rightarrow && a^2bc - ab^3-ac^3+b^2c^2 &= b^2c^2-2a^2bc+a^4 \\ \Rightarrow && 0 &= a^4+ab^3+ac^3-3a^2bc \\ &&&= a(a^3+b^3+c^3-3abc) \\ \underbrace{\Rightarrow}_{a \neq 0} && 0 &= a^3+b^3+c^3-3abc \\ &&&= (a+b+c)((a-b)^2+(b-c)^2+(c-a)^2) \end{align*} Therefore \(a+b+c = 0\). (Since otherwise \(a=b=c\) but \(ac \neq b^2\)). This means \(1\) is a root of our equations. Therefore, either \(k = 1\) or they have both roots in common, ie they are the same equation up to a scalar factor. ie \(b = la, c = lb, a= lc \Rightarrow l^3 = 1 \Rightarrow l = 1\). Therefore, they are the same equation.
Given that \(\cos A\), \(\cos B\) and \(\beta\) are non-zero, show that the equation \[ \alpha \sin(A-B) + \beta \cos(A+B) = \gamma \sin(A+B) \] reduces to the form \[ (\tan A-m)(\tan B-n)=0\,, \] where \(m\) and \(n\) are independent of \(A\) and \(B\), if and only if \(\alpha^2=\beta^2+\gamma^2\). Determine all values of \(x\), in the range \(0\le x <2\pi\), for which:
Solution: \begin{align*} && \alpha \sin(A-B) + \beta \cos (A + B) &= \gamma \sin(A+B) \\ \Leftrightarrow && \alpha \sin A \cos B - \alpha \cos A \sin B + \beta \cos A \cos B - \beta \sin A \sin B &= \gamma \sin A \cos B + \gamma \cos A \sin B \\ \Leftrightarrow && \alpha \tan A - \alpha \tan B + \beta - \beta \tan A \tan B &= \gamma \tan A + \gamma \tan B \\ \Leftrightarrow && \beta \tan A \tan B +(\gamma-\alpha) \tan A + (\gamma +\alpha)\tan B&=\beta \\ \Leftrightarrow && \tan A \tan B +\left (\frac{\gamma-\alpha}{\beta} \right) \tan A + \left (\frac{\gamma +\alpha}{\beta} \right)\tan B&=1\\ \Leftrightarrow && \left ( \tan A + \left ( \frac{\gamma+\alpha}{\beta} \right) \right) \left ( \tan B + \left ( \frac{\gamma-\alpha}{\beta} \right)\right) - \frac{\gamma^2 - \alpha^2}{\beta^2}&=1\\ \Leftrightarrow && \left ( \tan A + \left ( \frac{\gamma+\alpha}{\beta} \right) \right) \left ( \tan B + \left ( \frac{\gamma-\alpha}{\beta} \right)\right) &= \frac{\beta^2+\gamma^2-\alpha^2}{\beta^2}\\ \end{align*} Which has the desired form iff \(\beta^2+\gamma^2 = \alpha^2\).