Introduction to trig

Sine and cosine rule, graphs of trig functions, solving trig equations

Showing 1-13 of 13 problems
2019 Paper 1 Q6
D: 1500.0 B: 1518.2

In both parts of this question, \(x\) is real and \(0 < \theta < \pi\).

  1. By completing the square, find in terms of \(\theta\) the minimum value as \(x\) varies of $$9x^2 - 12x \cos \theta + 4.$$ Find also the maximum value as \(x\) varies of \(12x^2 \sin \theta - 9x^4\). Hence determine the values of \(x\) and \(\theta\) that satisfy the equation $$9x^4 + (9 - 12 \sin \theta)x^2 - 12x \cos \theta + 4 = 0.$$
  2. Sketch the curve $$y = \frac{x^2}{x - \theta},$$ where \(\theta\) is a constant. Deduce that either \(\frac{x^2}{x - \theta} \leq 0\) or \(\frac{x^2}{x - \theta} \geq 4\theta\). By considering the numerator and denominator separately, or otherwise, show that $$\frac{\sin^2 \theta \cos^2 x}{1 + \cos^2 \theta \sin^2 x} \leq 1.$$ Hence determine the values of \(x\) and \(\theta\) that satisfy the equation $$\frac{x^2}{4\theta(x - \theta)} = \frac{\sin^2 \theta \cos^2 x}{1 + \cos^2 \theta \sin^2 x}.$$

Show Solution
  1. \(\,\) \begin{align*} && y &= 9x^2 - 12x \cos \theta + 4 \\ &&&= (3x-2\cos \theta)^2+4-4\cos^2 \theta \\ &&&= (3x-2\cos \theta)^2 + 4 \sin^2 \theta \end{align*} Therefore the minimum is \(4\sin^2 \theta\) when \(x = \frac23 \cos \theta\). \begin{align*} && y &= 12x^2 \sin \theta - 9x^4 \\ &&&=4\sin^2 \theta -(3x^2-2\sin\theta)^2 \end{align*} Therefore the maximum is \(4\sin^2 \theta\) when \(x^2 = \frac23\sin \theta\) Therefore \begin{align*} && 0 &= 9x^4 + (9 - 12 \sin \theta)x^2 - 12x \cos \theta + 4 \\ && \underbrace{-9x^4+12x^2\sin \theta}_{\leq 4\sin^2 \theta } &= \underbrace{9x^2 - 12x \cos \theta + 4 }_{\geq 4 \sin^2 \theta} \end{align*} Therefore the equality cases must be achieved in both cases, ie \(x = \frac23 \cos \theta\) and \(x^2 = \frac23 \sin \theta\) \begin{align*} && x^2 &= \frac49\cos^2 \theta \\ &&&= \frac49(1-\sin^2 \theta) \\ &&&= \frac49(1-\frac94 x^2) \\ \Rightarrow && 2x^2 &= \frac49 \\ \Rightarrow && x &= \pm \frac{\sqrt{2}}3\\ \Rightarrow && \cos \theta &=\pm \frac32 \frac{\sqrt{2}}3 \\ &&&= \pm \frac{1}{\sqrt{2}} \\ \Rightarrow && \theta &= \frac{\pi}{4}, \frac{3\pi}{4} \\ \Rightarrow && (x, \theta) &= \left (\frac{\sqrt{2}}{3}, \frac{\pi}{4} \right), \left (-\frac{\sqrt{2}}{3}, \frac{3\pi}{4} \right) \end{align*}
  2. Sketching we obtain, noticing we can find the turning point by: \begin{align*} && \frac{x^2}{x-\theta} &= \lambda \\ \Leftrightarrow && x^2 - \lambda x +\theta \lambda &= 0 \\ \Leftrightarrow && 0 &\leq \Delta = \lambda^2 -4\lambda \theta \\ \Leftrightarrow && \lambda &\geq 4 \theta, \lambda \leq 0 \end{align*}
    TikZ diagram
    Notice that \(\sin^2 \theta \cos^2 x \leq 1\) and \(1 + cos^2 \theta \sin^2 x \geq 1\) and therefore we must have the inequality desired. \begin{align*} && \underbrace{\frac{x^2}{4\theta(x - \theta)}}_{\geq 1 \text{ or } \leq 0} &= \underbrace{\frac{\sin^2 \theta \cos^2 x}{1 + \cos^2 \theta \sin^2 x}}_{\in [0,1]} \\ \text{both}=0: && x = 0 &, \sin \theta = 0 \\ \text{both}=1: && x = 2\theta &, \sin^2 \theta = 1,\cos^2 x = 1 \\ && 1 &= \cos^2 2 \theta \\ &&&= (1-2 \sin^2 \theta)^2 \\ &&&= 1 \\ \Rightarrow && (x, \theta) &= \left(\frac{\pi}{2}, \pi\right) \end{align*}
2017 Paper 1 Q7
D: 1500.0 B: 1516.0

The triangle \(ABC\) has side lengths \(\left| BC \right| = a\), \(\left| CA \right| = b\) and \(\left| AB \right| = c\). Equilateral triangles \(BXC\), \(CYA\) and \(AZB\) are erected on the sides of the triangle \(ABC\), with \(X\) on the other side of \(BC\) from \(A\), and similarly for \(Y\) and \(Z\). Points \(L\), \(M\) and \(N\) are the centres of rotational symmetry of triangles \(BXC\), \(CY\!A\) and \(AZB\) respectively.

  1. Show that \(| CM| = \dfrac {\ b} {\sqrt3} \,\) and write down the corresponding expression for \(| CL|\).
  2. Use the cosine rule to show that \[ 6 \left| LM \right|^2 = a^2+b^2+c^2 + 4\sqrt3 \, \Delta \,, \] where \(\Delta\) is the area of triangle \(ABC\). Deduce that \(LMN\) is an equilateral triangle. Show further that the areas of triangles \(LMN\) and \(ABC\) are equal if and only if \[ a^2+b^2 +c^2 = 4\sqrt3 \, \Delta \,. \]
  3. Show that the conditions \[ (a -b)^2 = -2ab \big( 1 -\cos(C-60^\circ)\big) \,\] and \[ a^2+b^2 +c^2 = 4\sqrt3 \, \Delta \] are equivalent. Deduce that the areas of triangles \(LMN\) and \(ABC\) are equal if and only if \(ABC\) is equilateral.

Show Solution
TikZ diagram
  1. Consider the equilateral triangle \(CYA\), notice that \(YM\) is a vertical line of symmetry, and \(\angle ACM = 30^\circ\) therefore \(\frac{AC/2}{CM} = \cos 30^\circ \Rightarrow |CM| = \frac{b}{2} \cdot \frac{2}{\sqrt{3}} = \frac{b}{\sqrt{3}}\). Similarly \(|CL| = \frac{a}{\sqrt{3}}\)
  2. \(\,\) \begin{align*} && |LM|^2 &= |CM|^2 + |CL|^2 - 2 \cdot |CM| \cdot |CL| \cdot \cos \angle MCL \\ &&&= \frac{b^2}{3} + \frac{a^2}{3} - 2 \frac{ab}{3} \cos \left (\angle CMA + \angle CAB + \angle BCL \right) \\ &&&= \frac13 \left (b^2 + a^2 - 2ab \cos \left ( \frac{\pi}{3} + \angle CAB \right) \right) \\ &&&= \frac13 \left (b^2 + a^2 - ab \cos \left ( \angle CAB \right) + \sqrt{3}ab \sin \angle CAB \right) \\ &&&= \frac13 \left (b^2 + a^2 - ab \cos \angle CAB + 2\sqrt{3} \Delta\right) \\ &&&= \frac13 \left (b^2 + a^2 - ab \left (\frac{a^2+b^2-c^2}{2ab} \right) + 2\sqrt{3} \Delta\right) \\ &&&= \frac13 \left ( \frac12(a^2+b^2+c^2) + 2\sqrt{3}\Delta \right) \\ \Rightarrow && 6|LM|^2 &= a^2 + b^2 + c^2 + 4\sqrt{3} \Delta \end{align*} However, nothing in our reasoning here was special about \(LM\), therefore \(LN\) and \(MN\) also equal this value, and we find that the triangle is equilateral. The area of equilateral triangle [LMN] is \(\frac{\sqrt{3}}4 |LM|^2\), ie \begin{align*} &&& \text{areas are equal} \\ \Leftrightarrow && \Delta &= \frac{\sqrt{3}}4 |LM|^2 \\ &&&= \frac{\sqrt{3}}4 \frac{a^2+b^2+c^2+4\sqrt{3}\Delta}{6} \\ &&&= \frac{\sqrt{3}}{24} (a^2+b^2+c^2) + \frac12 \Delta \\ \Leftrightarrow && \Delta &= \frac{\sqrt{3}}{12}(a^2+b^2+c^2)\\ \Leftrightarrow && 4\sqrt{3}\Delta &=a^2+b^2+c^2\\ \end{align*}
  3. \(\,\) \begin{align*} && (a-b)^2 &= -2ab(1 - \cos(C - 60^{\circ})) \\ \Leftrightarrow && a^2+b^2 - 2ab &=-2ab + 2ab \cos(C - 60^{\circ}) \\ \Leftrightarrow && a^2+b^2 &= ab \cos C+\sqrt{3}ab\sin C \\ \Leftrightarrow && a^2+b^2 &= ab \frac{a^2+b^2-c^2}{2ab} + 2\sqrt{3} \Delta \\ \Leftrightarrow && a^2+b^2+c^2 &= 4\sqrt{3}\Delta \end{align*} Since the LHS is non-positive, and the RHS is positive, the only way they can be equal is if they are both \(0\), ie \(a=b\) and \(C = 60^{\circ}\) ie \(ABC\) is equilateral.
2015 Paper 1 Q3
D: 1484.0 B: 1516.0

A prison consists of a square courtyard of side \(b\) bounded by a perimeter wall and a square building of side \(a\) placed centrally within the courtyard. The sides of the building are parallel to the perimeter walls. Guards can stand either at the middle of a perimeter wall or in a corner of the courtyard. If the guards wish to see as great a length of the perimeter wall as possible, determine which of these positions is preferable. You should consider separately the cases \(b<3a\) and \(b>3a\,\).

Show Solution
TikZ diagram
TikZ diagram
TikZ diagram
The orange guard will always see \(2b+b-a = 3b-a\) The blue guard will see \(b + \frac{b(b-a)}{a} = \frac{b^2}{a}\) if \(b < 3a\) and \(3b + \frac{b(b-3a)}{(b-a)} = \frac{2b(2b-3a)}{b-a}\). Therefore the blue guard always sees more if \(b > 3a\). He sees more in the other case if \begin{align*} && \frac{b^2}{a} &> 3b - a \\ \Leftrightarrow && \frac{b^2}{a^2} &> 3\frac{b}{a} - 1 \\ \Leftrightarrow && x^2 - 3x + 1 &> 0\\ \Leftrightarrow && x > \frac{3 + \sqrt{5}}{2} \text{ or } x < \frac{3-\sqrt{5}}{2} \end{align*} Since \(b > a\) we must have \(b > \frac{3+\sqrt{5}}2 a\) There is an alternative interpretation which is that the orange guard is in the top left corner, ie
TikZ diagram
In this case the green guard will always see \(2b + \frac{2b(b-a)}{b+a} = \frac{4b^2}{b+a}\) Comparing \(\frac{4b^2}{b+a}\) with \(\frac{b^2}{a}\) we can see the former is larger if \(3a > b\). Comparing \(\frac{4b^2}{b+a}\) with $$
2015 Paper 2 Q2
D: 1600.0 B: 1484.0

In the triangle \(ABC\), angle \(BAC = \alpha\) and angle \(CBA= 2\alpha\), where \(2\alpha\) is acute, and \(BC= x\). Show that \(AB = (3-4 \sin^2\alpha)x\). The point \(D\) is the midpoint of \(AB\) and the point \(E\) is the foot of the perpendicular from \(C\) to \(AB\). Find an expression for \(DE\) in terms of \(x\). The point \(F\) lies on the perpendicular bisector of \(AB\) and is a distance \(x\) from \(C\). The points \(F\) and \(B\) lie on the same side of the line through \(A\) and \(C\). Show that the line \(FC\) trisects the angle \(ACB\).

Show Solution
TikZ diagram
Note that the sine rule gives us \begin{align*} && \frac{x}{\sin \alpha} &= \frac{AB}{\sin (180-3\alpha)} \\ \Rightarrow && AB &= x \frac{\sin 3\alpha}{\sin \alpha} \\ &&&= x \frac{\sin \alpha \cos 2\alpha + \cos \alpha \sin 2\alpha}{\sin \alpha} \\ &&&= x \left ( \frac{\sin \alpha (1-2\sin^2\alpha) + 2(1-\sin^2 \alpha)\sin \alpha}{\sin \alpha} \right) \\ &&&= x (3 - 4\sin^2 \alpha) \end{align*}
TikZ diagram
Note that \(AD = (\tfrac32 - 2 \sin^2\alpha)x\) and \(AE = (3-4\sin^2\alpha-\cos2\alpha)x\) so \begin{align*} DE &= (3-4\sin^2\alpha-\cos2\alpha)x - (\tfrac32 - 2 \sin^2\alpha)x \\ &= (\tfrac32 - 2\sin^2 \alpha - (1-2\sin^2 \alpha))x \\ &= \tfrac12x \end{align*}
TikZ diagram
Since \(DE = \frac12x\) if we drop the perpendicular from \(F\) to \(EC\) we have a \(30-60-90\) triangle. Therefore \(\angle FCE = 30^\circ\). Notice that \(\angle CEB = 90^{\circ} - 2\alpha\) and \(\angle ACB = 180^\circ - 3\alpha\), therefore \begin{align*} \angle ACF &= \angle ACB - \angle FCE - \angle ECB \\ &= (180^\circ - 3\alpha) - 30^\circ - (90^{\circ} - 2\alpha) \\ &= 60^\circ - \alpha \\ &= \frac13 \angle ACB \end{align*}
2014 Paper 2 Q1
D: 1600.0 B: 1500.0

In the triangle \(ABC\), the base \(AB\) is of length 1 unit and the angles at~\(A\) and~\(B\) are \(\alpha\) and~\(\beta\) respectively, where \(0<\alpha\le\beta\). The points \(P\) and~\(Q\) lie on the sides \(AC\) and \(BC\) respectively, with \(AP=PQ=QB=x\). The line \(PQ\) makes an angle of~\(\theta\) with the line through~\(P\) parallel to~\(AB\).

  1. Show that \(x\cos\theta = 1- x\cos\alpha - x\cos\beta\), and obtain an expression for \(x\sin\theta\) in terms of \(x\), \(\alpha\) and~\(\beta\). Hence show that \begin{equation} \label{eq:2*} \bigl(1+2\cos(\alpha+\beta)\bigr)x^2 - 2(\cos\alpha + \cos\beta)x + 1 = 0\,. \tag{\(*\)} \end{equation} Show that \((*)\) is also satisfied if \(P\) and \(Q\) lie on \(AC\) produced and \(BC\) produced, respectively. [By definition, \(P\) lies on \(AC\) produced if \(P\) lies on the line through \(A\) and~\(C\) and the points are in the order \(A\), \(C\), \(P\)\,.]
  2. State the condition on \(\alpha\) and \(\beta\) for \((*)\) to be linear in \(x\). If this condition does not hold (but the condition \(0<\alpha \le \beta\) still holds), show that \((*)\) has distinct real roots.
  3. Find the possible values of~\(x\) in the two cases (a) \(\alpha = \beta = 45^\circ\) and (b) \(\alpha = 30^\circ\), \(\beta = 90^\circ\), and illustrate each case with a sketch.

2012 Paper 2 Q6
D: 1600.0 B: 1528.8

A cyclic quadrilateral \(ABCD\) has sides \(AB\), \(BC\), \(CD\) and \(DA\) of lengths \(a\), \(b\), \(c\) and \(d\), respectively. The area of the quadrilateral is \(Q\), and angle \(DAB\) is \(\theta\). Find an expression for \(\cos\theta\) in terms of \(a\), \(b\), \(c\) and \(d\), and an expression for \(\sin\theta\) in terms of \(a\), \(b\), \(c\), \(d\) and \(Q\). Hence show that \[ 16Q^2 = 4(ad+bc)^2 - (a^2+d^2-b^2-c^2)^2 \,, \] and deduce that \[ Q^2 = (s-a)(s-b)(s-c)(s-d)\,, \] where \(s= \frac12(a+b+c+d)\). Deduce a formula for the area of a triangle with sides of length \(a\), \(b\) and \(c\).

Show Solution
TikZ diagram
\begin{align*} && BD^2 &= a^2+d^2 - 2ad \cos \theta \\ && BD^2 &= b^2+c^2-2bc \cos (\pi - \theta) \\ \Rightarrow && a^2+d^2 - 2ad \cos \theta &= b^2+c^2+2bc \cos \theta \\ \Rightarrow && 2(ad+bc)\cos \theta &= a^2+d^2-b^2-c^2 \\ \Rightarrow && \cos \theta &= \frac{a^2+d^2-b^2-c^2}{2(ad+bc)} \\ \\ && Q &= \frac12 ad \sin \theta + \frac12 bc \sin (\pi - \theta) \\ &&&= \frac12 (ad+bc) \sin \theta \\ \Rightarrow && \sin \theta &= \frac{2Q}{ad+bc} \\ \\ && 1 &= \sin^2 \theta + \cos^2 \theta \\ &&&= \frac{4Q^2}{(ad+bc)^2} + \frac{(a^2+d^2-b^2-c^2)^2}{4(ad+bc)^2} \\ \Rightarrow && 4(ad+bc)^2 &= 16Q^2 + (a^2+d^2-b^2-c^2)^2 \\ \Rightarrow && 16Q^2 &= 4(ad+bc)^2- (a^2+d^2-b^2-c^2)^2 \\ \Rightarrow && 16Q^2 &= (2ad+2bc - a^2-d^2+b^2+c^2)(2ad+2bc+a^2+d^2-b^2-c^2) \\ &&&= ((b+c)^2-(a-d)^2)((a+d)^2-(b-c)^2) \\ &&&= (b+c-a+d)(b+c+a-d)(a+d+b-c)(a+d-b+c) \\ \Rightarrow && Q^2 &= (s-a)(s-b)(s-c)(s-d) \end{align*} Since all triangles are cyclic, we can place \(D\) at the same point as \(A\) to obtain Heron's formula \(A = \sqrt{s(s-a)(s-b)(s-c)}\) where \(s = \frac12(a+b+c)\)
2010 Paper 2 Q6
D: 1600.0 B: 1516.0

Each edge of the tetrahedron \(ABCD\) has unit length. The face \(ABC\) is horizontal, and \(P\) is the point in \(ABC\) that is vertically below \(D\).

  1. Find the length of \(PD\).
  2. Show that the cosine of the angle between adjacent faces of the tetrahedron is \(1/3\).
  3. Find the radius of the largest sphere that can fit inside the tetrahedron.

Show Solution
  1. \(D\) must be above the centre (of any kind) of the equilateral triangle \(ABC\). Therefore it is a distance \(\frac23 \frac{\sqrt{3}}2 = \frac{\sqrt{3}}3\) from \(A\). \(D\) is \(1\) from \(A\), therefore by Pythagoras \(PD = \sqrt{1-\frac13} = \sqrt{\frac23}\)
  2. We can place \(D\) at \(\langle 0,0,\sqrt{\frac23}\rangle\) and \(A'\) (the midpoint of \(BC\)) at \(\langle-\frac{\sqrt{3}}{6},0,0 \rangle\) and we find: \begin{align*} && \cos \theta &= \frac{(\mathbf{a}'-\mathbf{d})\cdot \mathbf{a}'}{|\mathbf{a}'-\mathbf{d}|| \mathbf{a}'|} \\ &&&= \frac{|\mathbf{a}'|}{|\mathbf{a}'-\mathbf{d}|} \\ &&&= \frac{\frac{\sqrt{3}}{6}}{\sqrt{\frac23+\frac{3}{36}}} = \frac13 \end{align*}
  3. We have
    TikZ diagram
    And therefore we must have \(\tan \frac{\cos^{-1} \frac13}{2} = \frac{r}{\frac{\sqrt{3}}{6}}\) therefore \begin{align*} && r &= \frac{\sqrt{3}}{6} \tan \left (\frac{\cos^{-1} \frac13}{2} \right) \\ &&&= \frac{\sqrt{3}}6 \sqrt{\frac{1-\cos(\cos^{-1}\frac13)}{1+\cos(\cos^{-1}\frac13)}} \\ &&&= \frac{\sqrt{3}}6 \sqrt{\frac{\frac23}{\frac43}} \\ &&&= \frac{\sqrt{6}}{12} \end{align*}
2009 Paper 1 Q4
D: 1500.0 B: 1500.0

The sides of a triangle have lengths \(p-q\), \(p\) and \(p+q\), where \(p>q> 0\,\). The largest and smallest angles of the triangle are \(\alpha\) and \(\beta\), respectively. Show by means of the cosine rule that \[ 4(1-\cos\alpha)(1-\cos\beta) = \cos\alpha + \cos\beta \,. \] In the case \(\alpha = 2\beta\), show that \(\cos\beta=\frac34\) and hence find the ratio of the lengths of the sides of the triangle.

Show Solution
The largest angle will be opposite the side with length \(p+q\). Similarly the smallest angle will be opposite the side with length \(p-q\). The cosine rule tells us that: \begin{align*} && (p+q)^2 &= p^2 + (p-q)^2 - 2p(p-q) \cos \alpha \\ && 0 &= p(p-4q-2(p-q)\cos \alpha)\\ && 0 &= p(1-2\cos \alpha) + q(2\cos \alpha - 4)\\ \Rightarrow && \frac{p}{q} & = \frac{4-2 \cos \alpha}{1-2 \cos \alpha} \\ && (p-q)^2 &= p^2 + (p+q)^2 - 2p(p+q) \cos \beta \\ && 0 &= p(p+4q-2(p+q) \cos \beta) \\ && 0 &= p(1-2\cos \beta)+q(4-2\cos \beta) \\ \Rightarrow && \frac{p}{q} &= \frac{2\cos \beta - 4}{1-2\cos \beta} \\ \Rightarrow && \frac{4-2 \cos \alpha}{1-2 \cos \alpha} &= \frac{2\cos \beta - 4}{1-2\cos \beta} \\ \Rightarrow && (2-\cos \alpha)(1-2\cos \beta) &= (\cos \beta - 2)(1 - 2 \cos \alpha) \\ \Rightarrow && 2 - \cos \alpha -4\cos \beta+2\cos \alpha \cos \beta &= \cos \beta - 2-2\cos \alpha \cos \beta + 4 \cos \alpha \\ \Rightarrow && 4-4\cos \alpha - 4\cos \beta+4\cos \alpha\cos \beta &= \cos \alpha + \cos \beta \\ \Rightarrow && 4(1-\cos \alpha)(1-\cos \beta) &= \cos \alpha + \cos \beta \end{align*} If \(\alpha = 2 \beta\), and let \(c = \cos \beta\) \begin{align*} && 4 (1- \cos 2 \beta)(1-\cos \beta) &= \cos 2 \beta + \cos \beta \\ \Rightarrow && 4(1-(2c^2-1))(1-c) &= 2c^2-1+c\\ \Rightarrow && 8(1+c)(1-c)^2 &= (2c-1)(c+1) \\ \Rightarrow && 0 &= (c+1)(8(1-c)^2-(2c-1)) \\ &&&= (c+1)(8c^2-18c+9) \\ &&&= (c+1)(4c-3)(2c-3) \\ \end{align*} Therefore \(c = -1, \frac32, \frac34\). Clearly \(\cos \beta \neq -1, \frac32\), since they are not valid angles in a triangle (or valid values of \(\cos \beta\)). \(\frac{p}{q} = \frac{2 \cdot \frac34-4 }{1 - 2\cdot \frac34} = \frac{3-8}{2-3} = 5\) so \(4:5:6\)
2009 Paper 1 Q8
D: 1500.0 B: 1484.0

  1. The equation of the circle \(C\) is \[ (x-2t)^2 +(y-t)^2 =t^2, \] where \(t\) is a positive number. Show that \(C\) touches the line \(y=0\,\). Let \(\alpha\) be the acute angle between the \(x\)-axis and the line joining the origin to the centre of \(C\). Show that \(\tan2\alpha=\frac43\) and deduce that \(C\) touches the line \(3y=4x\,\).
  2. Find the equation of the incircle of the triangle formed by the lines \(y=0\), \(3y=4x\) and \(4y+3x=15\,\). Note: The incircle of a triangle is the circle, lying totally inside the triangle, that touches all three sides.

Show Solution
  1. This is a circle centre \((2t,t)\) with radius \(t\). Therefore it is exactly \(t\) away from the line \(y = 0\) so just touches that line. Not that \(\tan \alpha = \frac{t}{2t} = \frac12\) so \(\tan 2\alpha = \frac{2\tan \alpha}{1-\tan^2\alpha} = \frac{1}{1-\frac14} = \frac43\). Therefore the line \(y = \frac43x\) or \(3y = 4x\) is tangent to \(C\).
  2. Note that \(3y=4x\) and \(4y+3x=15\) are perpendicular, so this is a right-angled triangle with incenter \((2t,t)\) for some \(t\) and hypotenuse \(15\) We can find the third coordinate when \(3y-4x = 0\) and \(4y+3x = 15\) meet, ie \((\frac{9}{5}, \frac{12}5)\) The incentre lies on the bisector of the right angle at this point, which is the line through \((\frac{9}{5}, \frac{12}5)\) and \((\frac{15}{2}, 0)\), so \begin{align*} && \frac{2t-\frac{12}{5}}{t - \frac{9}{5}} &= \frac{-\frac{12}{5}}{\frac{15}2-\frac95} \\ \Rightarrow && \frac{10t-12}{5t-9} &= \frac{-24}{57} = -\frac{8}{19} \\ \Rightarrow && 190t - 12 \cdot 19 &= -40t + 72 \\ \Rightarrow && t &= 2 \end{align*} Therefore the center is \((4, 2)\) and the equation is \((x-4)^2+(y-2)^2=2^2\)
2009 Paper 2 Q3
D: 1600.0 B: 1500.0

Prove that \[ \tan \left ( \tfrac14 \pi -\tfrac12 x \right)\equiv \sec x -\tan x\,. \tag{\(*\)} \]

  1. Use \((*)\) to find the value of \(\tan\frac18\pi\,\). Hence show that \[ \tan \tfrac{11}{24} \pi = \frac{\sqrt3 + \sqrt2 -1}{\sqrt3 -\sqrt6+1}\;. \]
  2. Show that \[ \frac{\sqrt3 + \sqrt2 -1}{\sqrt3 -\sqrt6+1}= 2+\sqrt2+\sqrt3+\sqrt6\,. \]
  3. Use \((*)\) to show that \[ \tan \tfrac1{48}\pi = \sqrt{16+10\sqrt2+8\sqrt3 +6\sqrt6 \ }-2-\sqrt2-\sqrt3-\sqrt6\,. \]

Show Solution
\begin{align*} && \tan \left ( \tfrac14 \pi -\tfrac12 x \right) &\equiv \frac{\tan \tfrac{\pi}{4} - \tan \tfrac12 x}{1 + \tan \tfrac{\pi}{4} \tan \tfrac12 x} \\ &&&= \frac{1-\tan \tfrac12 x}{1+\tan \tfrac12 x} \\ \\ && \sec x - \tan x &= \frac{1+t^2}{1-t^2} - \frac{2t}{1-t^2} \\ &&&= \frac{(1-t)^2}{(1-t)(1+t)} \\ &&&= \frac{1-t}{1+t} \end{align*} Therefore both sides are equal to the same thing.
  1. \(\tan \tfrac18 \pi = \tan(\tfrac14 \pi - \tfrac12 \tfrac14\pi) = \sec \tfrac14 \pi - \tan \tfrac14 \pi = \sqrt{2} - 1\) \begin{align*} && \tan \tfrac{11}{24} \pi &= \tan (\tfrac13 \pi +\tfrac18 \pi) \\ &&&= \frac{\tan \tfrac13 \pi +\tan \tfrac18 \pi}{1-\tan \tfrac13 \pi \tan \tfrac18 \pi} \\ &&&= \frac{\sqrt{3} + \sqrt{2} - 1}{1 - \sqrt{3}(\sqrt{2}-1)} \\ &&&= \frac{\sqrt{3} + \sqrt{2} - 1}{1 +\sqrt{3}-\sqrt{6}} \\ \end{align*}
  2. \(\,\) \begin{align*} && (\sqrt{3}-\sqrt{6}+1)(2+\sqrt{2}+\sqrt{3}+\sqrt{6}) &= (2\sqrt{3}+\sqrt{6}+3+3\sqrt{2}) + \\ &&&\quad+(-2\sqrt{6}-2\sqrt{3}-3\sqrt{2}-6) + \\ &&&\quad+(2+\sqrt{2}+\sqrt{3}+\sqrt{6}) \\ &&&= \sqrt{3}+\sqrt{2}-1 \end{align*}
  3. \(\,\) \begin{align*} && \tan \tfrac{1}{48} \pi &= \tan (\tfrac14\pi - \tfrac{11}{48} \pi) \\ &&&= \sec \tfrac{11}{24} \pi - \tan \tfrac{11}{24} \pi \\ &&&= \sqrt{1+\tan^2 \tfrac{11}{24}\pi} - \tan \tfrac{11}{24} \pi \\ &&&= \sqrt{1 + (2+\sqrt{2}+\sqrt{3}+\sqrt{6})^2} - (2+\sqrt{2}+\sqrt{3}+\sqrt{6}) \\ &&&= \sqrt{16+10\sqrt{2} + 8\sqrt{3}+6\sqrt{6}} - 2 - \sqrt2 - \sqrt3-\sqrt6 \end{align*}
2006 Paper 1 Q8
D: 1500.0 B: 1484.0

{\it Note that the volume of a tetrahedron is equal to \(\frac1 3\) \(\times\) the area of the base \(\times\) the height.} The points \(O\), \(A\), \(B\) and \(C\) have coordinates \((0,0,0)\), \((a,0,0)\), \((0,b,0)\) and \((0,0,c)\), respectively, where \(a\), \(b\) and \(c\) are positive.

  1. Find, in terms of \(a\), \(b\) and \(c\), the volume of the tetrahedron \(OABC\).
  2. Let angle \(ACB = \theta\). Show that \[ \cos\theta = \frac {c^2} { { \sqrt{\vphantom{ \dot b} (a^2+c^2)(b^2+c^2)} } ^{\vphantom A} \ } \] and find, in terms of \(a\), \(b\) and \(c\), the area of triangle \(ABC\). % is %\(\displaystyle \tfrac12 \sqrt{ \vphantom{\dot A } a^2b^2 +b^2c^2 + c^2 a^2 \;} \;\).
Hence show that \(d\), the perpendicular distance of the origin from the triangle \(ABC\), satisfies \[ \frac 1{d^2} = \frac 1 {a^2} + \frac 1 {b^2} + \frac 1 {c^2} \,. \]

2002 Paper 3 Q7
D: 1700.0 B: 1484.0

Given that \(\alpha\) and \(\beta\) are acute angles, show that \(\alpha + \beta = \tfrac{1}{2}\pi\) if and only if \(\cos^2 \alpha + \cos^2 \beta = 1\). In the \(x\)--\(y\) plane, the point \(A\) has coordinates \((0,s)\) and the point \(C\) has coordinates \((s,0)\), where \(s>0\). The point \(B\) lies in the first quadrant (\(x>0\), \(y>0\)). The lengths of \(AB\), \(OB\) and \(CB\) are respectively \(a\), \(b\) and \(c\). Show that \[ (s^2 +b^2 - a^2)^2 + (s^2 +b^2 -c^2)^2 = 4s^2b^2 \] and hence that \[ (2s^2 -a^2-c^2)^2 + (2b^2 -a^2-c^2)^2 =4a^2c^2\;. \] Deduce that $$ \l a - c \r^2 \le 2b^2 \le \l a + c \r^2\;. $$ %Show, %by considering the case \(a=1+\surd2\,\), \(b=c=1\,\), % that the condition \(\l \ast \r\,\) %is not sufficient to ensure that \(B\) lies in the first quadrant.

2000 Paper 1 Q5
D: 1500.0 B: 1484.0

Arthur and Bertha stand at a point \(O\) on an inclined plane. The steepest line in the plane through \(O\) makes an angle \(\theta\) with the horizontal. Arthur walks uphill at a steady pace in a straight line which makes an angle \(\alpha\) with the steepest line. Bertha walks uphill at the same speed in a straight line which makes an angle \(\beta\) with the steepest line (and is on the same side of the steepest line as Arthur). Show that, when Arthur has walked a distance \(d\), the distance between Arthur and Bertha is \(2d \vert\sin\frac12(\alpha-\beta)\vert\). Show also that, if \(\alpha\ne\beta\), the line joining Arthur and Bertha makes an angle \(\phi\) with the vertical, where \[ \cos\phi = \sin\theta \sin \frac12(\alpha+\beta). \]