Problems

Filters
Clear Filters

39 problems found

2008 Paper 2 Q6
D: 1600.0 B: 1484.0
trigonometry compound angle formulae period of function sum-to-product roots of trigonometric equations curve sketching trigonometric identities

A curve has the equation \(y=\f(x)\), where \[ \f(x) = \cos \Big( 2x+ \frac \pi 3\Big) + \sin \Big ( \frac{3x}2 - \frac \pi 4\Big). \]

  1. Find the period of \(\f(x)\).
  2. Determine all values of \(x\) in the interval \(-\pi\le x \le \pi\) for which \(\f(x)=0\). Find a value of \(x\) in this interval at which the curve touches the \(x\)-axis without crossing it.
  3. Find the value or values of \(x\) in the interval \(0\le x \le 2\pi\) for which \(\f(x)=2\,\).

Solution: \begin{align*} && f(x) &= \cos \left( 2x+ \frac \pi 3\right) + \sin \left ( \frac{3x}2 - \frac\pi 4\right) \\ &&&= \cos \left( 2x+ \frac \pi 3\right) + \cos\left (\frac{\pi}{2} - \left ( \frac{3x}2 - \frac\pi 4\right) \right)\\ &&&= \cos \left( 2x+ \frac \pi 3\right) + \cos\left (\frac{3\pi}{4} - \frac{3x}2 \right)\\ &&&= 2 \cos \left (\frac{2x+ \frac \pi 3+\frac{3\pi}{4} - \frac{3x}2}{2} \right) \cos \left ( \frac{\left (2x+ \frac \pi 3 \right) - \left (\frac{3\pi}{4} - \frac{3x}2 \right)}{2} \right)\\ &&&= 2 \cos \left (\frac{\frac{x}{2}+ \frac {13\pi}{12}}{2} \right) \cos \left ( \frac{\frac{7x}{2}- \frac {5\pi}{12}}{2} \right)\\ &&&= 2 \cos \left (\frac{x}{4}+ \frac {13\pi}{24} \right) \cos \left ( \frac{7x}{4}- \frac {5\pi}{24} \right)\\ \end{align*}

  1. The period of \(f\) will be the LCM of \(\frac{2\pi}{\pi}\) and \(\frac{2\pi}{\frac32} = \frac{4\pi}{3}\) which is \(4\pi\). (This is also clear from the factorised form).
  2. \(f(x) = 0\) means one of those two factors is zero, ie \begin{align*} \text{first factor}: && 0 &= \cos \left (\frac{x}{4}+ \frac {13\pi}{24} \right) \\ &&n\pi + \frac{\pi}{2}&= \frac{x}{4}+ \frac {13\pi}{24} \\ \Rightarrow && x &= 4n\pi - \frac{\pi}{6} \\ \Rightarrow && x &= -\frac{\pi}{6} \\ \\ \text{second factor}: && 0 &= \cos \left ( \frac{7x}{4}- \frac {5\pi}{24} \right) \\ && n\pi + \frac{\pi}{2} &= \frac{7x}{4}- \frac {5\pi}{24} \\ \Rightarrow && 7x &= 4n\pi + \frac{17}{6}\pi \\ \Rightarrow && x &= \frac{4n}7\pi + \frac{17}{42}\pi \\ \Rightarrow && x &= -\frac{31}{42} \pi, -\frac16\pi, \frac{17}{42}\pi, \frac{41}{42}\pi \end{align*} Therefore all solutions are \(-\frac{31}{42} \pi, -\frac16\pi, \frac{17}{42}\pi, \frac{41}{42}\pi\) We can see that \(-\frac{\pi}{6}\) is a repeated root, therefore it touches the axis and does not cross.
  3. \(f(x) = 2\) requires both factors to be \(1\) or \(-1\). \begin{align*} \text{first factor}: && \pm1 &= \cos \left (\frac{x}{4}+ \frac {13\pi}{24} \right) \\ &&n\pi &= \frac{x}{4}+ \frac {13\pi}{24} \\ \Rightarrow && x &= 4n\pi - \frac{13\pi}{6} \\ \Rightarrow && x &= \frac{11}{6}\pi \\ \end{align*} We only need to test this value, where it's \(-1\), so we look at \( \cos \left ( \frac{77\pi}{24}- \frac {5\pi}{24} \right) = \cos (3\pi) = -1\), so the only value is \(\frac{11}{6}\pi\)

View
2007 Paper 3 Q1
D: 1700.0 B: 1500.0
roots of polynomials symmetric functions Vieta's formulas trigonometric identities tan addition formula quartic t-substitution double angle formula

In this question, do not consider the special cases in which the denominators of any of your expressions are zero. Express \(\tan(\theta_1+\theta_2+\theta_3+\theta_4)\) in terms of \(t_i\), where \(t_1=\tan\theta_1\,\), etc. Given that \(\tan\theta_1\), \(\tan\theta_2\), \(\tan\theta_3\) and \(\tan\theta_4\) are the four roots of the equation \[at^4+bt^3+ct^2+dt+e=0 \] (where \(a\ne0\)), find an expression in terms of \(a\), \(b\), \(c\), \(d\) and \(e\) for \(\tan(\theta_1+\theta_2+\theta_3+\theta_4)\). The four real numbers \(\theta_1\), \(\theta_2\), \(\theta_3\) and \(\theta_4\) lie in the range \(0\le \theta_i<2\pi\) and satisfy the equation \[ p\cos2\theta+\cos(\theta-\alpha)+p=0\,,\] where \(p\) and \(\alpha\) are independent of \(\theta\). Show that \(\theta_1+\theta_2+\theta_3+\theta_4=n\pi\) for some integer \(n\).

Solution: \begin{align*} \tan(\theta_1 + \theta_2 + \theta_3 + \theta_4) &= \frac{\tan(\theta_1 + \theta_2) + \tan(\theta_3 + \theta_4)}{1 - \tan(\theta_1 +\theta_2)\tan(\theta_3+\theta_4)} \\ &= \frac{\frac{t_1+t_2}{1-t_1t_2}+\frac{t_3+t_4}{1-t_3t_4}}{1-\frac{t_1+t_2}{1-t_1t_2}\frac{t_3+t_4}{1-t_3t_4}} \\ &= \frac{(t_1+t_2)(1-t_3t_4)+(t_3+t_4)(1-t_1t_2)}{(1-t_1t_2)(1-t_3t_4)-(t_1+t_2)(t_3+t_4)} \\ &= \frac{t_1 +t_2+t_3+t_4 - (t_1t_2t_3+t_1t_2t_4+t_1t_3t_4+t_2t_3t_4)}{1-t_1t_2-t_1t_3-t_1t_4-t_2t_3-t_2t_4-t_3t_4} \end{align*} If \(t_1, t_2, t_3, t_4\) are the roots of \(at^4+bt^3+ct^2+dt+e = 0\), then \(t_1+t_2+t_3+t_4 = -\frac{b}{a}, t_1t_2+t_1t_3+t_1t_4+t_2t_3+t_2t_4+t_3t_4 = \frac{c}{a}, t_1t_2t_3+t_1t_2t_4+t_1t_3t_4+t_2t_3t_4 = -\frac{d}{a}\), therefore the expression is: \begin{align*} \tan(\theta_1 + \theta_2 + \theta_3 + \theta_4) &= \frac{-\frac{b}{a}+\frac{d}{a}}{1 - \frac{c}{a}} \\ &= \frac{d-b}{a-c} \end{align*} \begin{align*} &&0 &= p \cos 2\theta + \cos (\theta - \alpha) + p \\ &&&= p (2\cos^2 \theta -1) + \cos \theta \cos \alpha - \sin \theta \sin \alpha + p \\ &&&= 2p \cos^2 \theta + \cos \theta \cos \alpha - \sin \theta \sin \alpha\\ \Rightarrow && 0 &=2p \cos \theta + \cos \alpha - \tan \theta \sin \alpha \\ \Rightarrow && -2p \cos \theta&= \cos \alpha - \tan \theta \sin \alpha \\ \Rightarrow && 4p^2 \cos^2 \theta &= \cos^2 \alpha - 2 \sin \alpha \cos \alpha \tan \theta + \sin^2 \alpha \tan^2 \theta \\ && 4p^2 \frac{1}{1 + \tan^2 \theta} &= \cos^2 \alpha - \sin 2\alpha \tan \theta + \sin^2 \alpha \tan^2 \theta \\ \Rightarrow && 4p^2 &= \cos^2 \alpha - \sin 2\alpha t+t^2-\sin2\alpha t^3+\sin^2 \alpha t^4 \\ \Rightarrow && \tan (\theta_1+\theta_2 + \theta_3+ \theta_4) &= \frac{0}{\sin^2 \alpha - 1} \\ &&&= 0 \\ \Rightarrow && \theta_1 + \theta_2 + \theta_3 + \theta_4 &= n\pi \end{align*}

View
2006 Paper 3 Q3
D: 1700.0 B: 1500.0
Taylor series trigonometric power series cot tan cosec trigonometric identities series coefficients

  1. Let \[ \tan x = \sum\limits_{n=0}^\infty a_n x^n \text{ and } \cot x = \dfrac 1 x +\sum\limits_{n=0}^\infty b_nx^n \] for \(0< x < \frac12\pi\,\). Explain why \(a_n=0\) for even \(n\). Prove the identity \[ \cot x - \tan x \equiv 2 \cot 2x\, \] and show that \[a_{n} = (1-2^{n+1})b_n\,.\]
  2. Let $ \displaystyle {\rm cosec}\, x = \frac1x +\sum\limits _{n=0}^\infty c_n x^n\,$ for \(0< x < \frac12\pi\,\). By considering \(\cot x + \tan x\), or otherwise, show that \[ c_n = (2^{-n} -1)b_n \,. \]
  3. Show that \[ \left(1+x{ \sum\limits_{n=0}^\infty} b_n x^n \right)^2 +x^2 = \left(1+x{ \sum\limits_{n=0} ^\infty} c_n x^n \right)^2\,. \] Deduce from this and the previous results that \(a_1=1\), and find \(a_3\).

Solution:

  1. Since \(\tan (-x) = -\tan x\), \(\tan\) is an odd function, and in particular all it's even coefficients are zero. \begin{align*} && 2 \cot 2x &\equiv \frac{2 cos 2x}{\sin 2 x} \\ &&&\equiv \frac{2(\cos^2 x- \sin^2 x)}{2 \sin x \cos x} \\ &&&\equiv \frac{\cos x}{\sin x} - \frac{\sin x}{ \cos x} \\ &&&\equiv \cot x - \tan x \end{align*} Therefore \begin{align*} && \underbrace{\frac1x + \sum_{n=0}^\infty b_nx^n}_{\cot x} - \underbrace{\sum_{n=0}^\infty a_n x^n}_{\tan x} &= 2 \left (\underbrace{\frac{1}{2x} + \sum_{n=0}^\infty b_n(2x)^n}_{\cot 2x} \right) \\ \Rightarrow && \sum_{n=0}^\infty a_n x^n &= \sum_{n=0}^\infty b_nx^n - 2\sum_{n=0}^\infty b_n(2x)^n \\ &&&= \sum_{n=0}^{\infty}b_n(1-2^{n+1})x^n \\ [x^n]: && a_n &= (1-2^{n+1})b_n \end{align*}
  2. \(\,\) \begin{align*} && \cot x + \tan x &= \frac{\cos x}{\sin x} + \frac{\sin x}{\cos x} \\ &&&= \frac{1}{\sin x \cos x} \\ &&&=2\cosec 2x \\ \\ \Rightarrow && \underbrace{\frac1x + \sum_{n=0}^\infty b_nx^n}_{\cot x} + \underbrace{\sum_{n=0}^\infty a_n x^n}_{\tan x} &= 2\left (\underbrace{ \frac1{2x} +\sum\limits _{n=0}^\infty c_n (2x)^n}_{\cosec 2x} \right) \\ \Rightarrow && \sum_{n=0}^\infty 2^{n+1}c_n x^n &= \sum_{n=0}^{\infty}(a_n+b_n)x^n \\ &&&= \sum_{n=0}^{\infty}\left((1-2^{n+1})b_n+ b_n\right)x^n \\ &&&= \sum_{n=0}^{\infty}\left(2-2^{n+1}\right)b_nx^n \\ [x^n]: && c_n &= (2^{-n}-1)b_n \end{align*}
  3. \(\,\) \begin{align*} && \cot^2 x + 1 &= \cosec^2 x \\ \Rightarrow && x^2 \cot^2 x + x^2 &= x^2 \cosec^2 x \\ \Rightarrow && x^2 \left ( \underbrace{\frac1x + \sum_{n=0}^\infty b_nx^n}_{\cot x} \right)^2 + x^2 &= x^2 \left (\underbrace{ \frac1{x} +\sum\limits _{n=0}^\infty c_n x^n}_{\cosec x} \right)^2 \\ \Rightarrow && \left ( 1 + x\sum_{n=0}^\infty b_nx^{n} \right)^2 + x^2 &= \left ( 1 +x\sum\limits _{n=0}^\infty c_n x^{n} \right)^2 \\ \\ \Rightarrow && \left ( 1 + x(b_1x + b_3 x^3 + \cdots) \right)^2 + x^2 &= \left ( 1 + x(c_1x + c_3 x^3 + \cdots) \right)^2 \\ \Rightarrow && 1 + (1+2b_1)x^2+(2b_3+b_1^2)x^4 + \cdots &= 1 + 2c_1x^2 + (2c_3+c_1^2)x^4 + \cdots \\ \Rightarrow && 1 + 2b_1 &= 2(2^{-1}-1)b_1 \\ \Rightarrow && b_1 &= -\frac13 \\ \Rightarrow && a_1 &= (1-2^{2})(-\tfrac13) = 1 \\ && c_1 &= \frac16\\ \Rightarrow && 2b_3+\frac19&= 2c_3+\frac1{36} \\ \Rightarrow && 2b_3 -2(2^{-3}-1)b_3 &= -\frac{1}{12} \\ \Rightarrow && \frac{15}{4}b_3 &= -\frac{1}{12} \\ \Rightarrow && b_3 &= -\frac{1}{45} \\ \Rightarrow && a_3 &= -(1-2^4)\frac{1}{45} = \frac13 \end{align*}

View
2002 Paper 2 Q9
D: 1600.0 B: 1500.0
projectiles parabolic trajectory trigonometric identities optimisation small increments perturbation analysis simultaneous equations inequality

A particle is projected from a point \(O\) on a horizontal plane with speed \(V\) and at an angle of elevation \(\alpha\). The vertical plane in which the motion takes place is perpendicular to two vertical walls, both of height \(h\), at distances \(a\) and \(b\) from \(O\). Given that the particle just passes over the walls, find \(\tan\alpha\) in terms of \(a\), \(b\) and \(h\) and show that \[ \frac{2V^2} g = \frac {ab} h +\frac{ (a+b)^2 h}{ab} \;. \] The heights of the walls are now increased by the same small positive amount \(\delta h\,\). A second particle is projected so that it just passes over both walls, and the new angle and speed of projection are \(\alpha +\delta \alpha \) and \(V+\delta V\), respectively. Show that \[ \sec^2 \alpha \, \delta \alpha \approx \frac {a+b}{ab}\,\delta h \;, \] and deduce that \(\delta \alpha >0\,\). Show also that \(\delta V\) is positive if \(h> ab/(a+b)\) and negative if \(h

View
2001 Paper 2 Q4
D: 1600.0 B: 1484.0
trigonometry sin nx quadratic in cos roots of equations discriminant factorization proof trigonometric identities

Let $$ \f(x) = P \, {\sin x} + Q\, {\sin 2x} + R\, {\sin 3x} \;. $$ Show that if \(Q^2 < 4R(P-R)\), then the only values of \(x\) for which \(\f(x) = 0\) are given by \(x=m\pi\), where \(m\) is an integer. \newline [You may assume that \(\sin 3x = \sin x(4\cos^2 x -1)\).] Now let $$ \g(x) = {\sin 2nx} + {\sin 4nx} - {\sin 6nx}, $$ where \(n\) is a positive integer and \(0 < x < \frac{1}{2}\pi \). Find an expression for the largest root of the equation \(\g(x)=0\), distinguishing between the cases where \(n\) is even and \(n\) is odd.

View
2000 Paper 2 Q6
D: 1600.0 B: 1585.4
integration Weierstrass substitution t-substitution trigonometric trigonometric identities definite integral half-angle substitution

Show that \[ \sin\theta = \frac {2t}{1+t^2}, \ \ \ \cos\theta = \frac{1-t^2}{1+t^2}, \ \ \ \frac{1+\cos\theta}{\sin\theta} = \tan (\tfrac{1}{2}\pi-\tfrac{1}{2}\theta), \] where \(t =\tan\frac{1}{2}\theta\). Use the substitution \(t =\tan\frac{1}{2}\theta\) to show that, for \(0<\alpha<\frac{1}{2}\pi\), \[ \int_0^{\frac{1}{2}\pi} {1 \over {1 + \cos\alpha \sin \theta}} \,\d\theta =\frac{\alpha}{\sin\alpha}\,, \] and deduce a similar result for \[ \int_0^{\frac{1}{2}\pi} {1 \over {1 + \sin\alpha \cos \theta}} \,\d\theta \,. \]

Solution: \begin{align*} && \frac{2t}{1+t^2} &= \frac{2 \sin \tfrac12 \theta \cos\tfrac12 \theta }{\cos^2 \tfrac12 \theta + \sin^2 \tfrac12 \theta} \\ &&&= \frac{\sin \theta}{1} = \sin \theta \\ \\ && \frac{1-t^2}{1+t^2} &= \frac{\cos^2 \tfrac12 \theta - \sin^2 \tfrac12 \theta}{\cos^2 \tfrac12 \theta + \sin^2 \tfrac12 \theta} \\ &&&= \frac{\cos \theta }{1} = \cos \theta \\ \\ && \tan(\tfrac12 \pi - \tfrac12 \theta) &= \frac{1}{t} \\ && \frac{1+\cos \theta}{\sin \theta} &= \frac{1 + \frac{1-t^2}{1+t^2}}{\frac{2t}{1+t^2}} \\ &&&= \frac{2}{2t} = \frac1t = \tan(\tfrac12\pi - \tfrac12 \theta) \end{align*} Notice also that \(\frac{\d t}{\d \theta} = \tfrac12 \sec^2 \tfrac12 \theta = \tfrac12(1 + t^2)\) so \begin{align*} && I &= \int_0^{\frac12 \pi} \frac{1}{1 + \cos \alpha \sin \theta} \d \theta \\ t = \tan \tfrac12 \theta, \d \theta = \frac{2}{1+t^2} \d t: &&&= \int_{0}^{1} \frac{1}{1 + \cos \alpha \frac{2t}{1+t^2}}\frac{2}{1+t^2} \d t \\ &&&= \int_0^1 \frac{2}{1+t^2 + 2\cos \alpha t} \d t \\ &&&= \int_0^1 \frac{2}{(t + \cos \alpha)^2+\sin^2 \alpha} \d t \\ &&&= \left [ \frac{2}{\sin \alpha} \tan^{-1} \left ( \frac{t+ \cos \alpha}{\sin \alpha} \right) \right]_0^1 \\ &&&= \frac{2}{\sin \alpha} \left ( \tan^{-1} \left ( \frac{1+ \cos \alpha}{\sin \alpha} \right) - \tan^{-1} \left ( \frac{ \cos \alpha}{\sin \alpha} \right) \right) \\ &&&= \frac{2}{\sin \alpha} \left ( \tan^{-1} \left (\tan (\tfrac12 \pi - \tfrac12 \alpha \right) - \tan^{-1} \left (\tan(\tfrac12\pi - \alpha )\right) \right) \\ &&&= \frac{2}{\sin \alpha} \left ( \tfrac12 \pi - \tfrac12 \alpha - \tfrac12 \pi + \alpha \right) \\ &&&= \frac{\alpha}{\sin \alpha} \end{align*} \begin{align*} && J &= \int_0^{\tfrac12 \pi} \frac{1}{1 + \sin \alpha \cos \theta} \d \theta \\ &&&= \int_0^{\tfrac12 \pi} \frac{1}{1 + \cos (\tfrac12 \pi - \alpha) \sin \theta} \d \theta \\ &&&= \frac{\tfrac12 \pi - \alpha}{\cos \alpha} \end{align*}

View
2000 Paper 3 Q2
D: 1700.0 B: 1484.2
integration trigonometric substitution definite integral square root generalisation trigonometric identities

Use the substitution \(x = 2-\cos \theta \) to evaluate the integral $$ \int_{3/2}^2 \left(x - 1 \over 3 - x\right)^{\!\frac12}\! \d x. $$ Show that, for \(a < b\), $$ \int_p^q \left( x - a \over b - x\right)^{\!\frac12} \!\d x = \frac{(b-a)(\pi +3{\surd3} -6)}{12}, $$ where \(p= {(3a+b)/4}\) and \(q={(a+b)/2}\).

View
1998 Paper 1 Q3
D: 1500.0 B: 1500.0
true-false logarithms exponentials trigonometric identities polynomial approximation inequality AM-GM inequality proof

Which of the following statements are true and which are false? Justify your answers.

  1. \(a^{\ln b}=b^{\ln a}\) for all \(a,b>0\).
  2. \(\cos(\sin\theta)=\sin(\cos\theta)\) for all real \(\theta\).
  3. There exists a polynomial \(\mathrm{P}\) such that \(|\mathrm{P}(\theta)-\cos\theta|\leqslant 10^{-6}\) for all real \(\theta\).
  4. \(x^{4}+3+x^{-4}\geqslant 5\) for all \(x>0\).

Solution:

  1. True. \begin{align*} && \ln a \cdot \ln b &= \ln b \cdot \ln a \\ \Leftrightarrow && \exp ( \ln a \cdot \ln b) &= \exp ( \ln b \cdot \ln a) \\ \Leftrightarrow && \exp ( \ln a )^{\ln b} &= \exp ( \ln b )^{\ln a} \\ \Leftrightarrow && a^{\ln b} &= b^{\ln a} \\ \end{align*}
  2. False. Consider \(\theta = 0\). We'd need \(\cos 0 = 1 = \sin 1\), but \(0 < 1 < \frac{\pi}{2}\) so \(\sin 1 \neq 1\)
  3. False. If the polynomial has positive degree, then as \(n \to \infty\), \(\P(x) \to \pm \infty\), in particular it must be well outside the interval \([-1,1]\). Therefore it can't be within \(10^{-6}\) of \(\cos \theta\) which is confined to that interval. The only polynomial which is restricted to that range are constants, but then \(|\cos 0 - c| \leq 10^{-6}\) and \(|\cos \pi - c| \leq 10^{-6}\) \(2 = |1-(-1)| \leq |1-c| + |-1-c| \leq 2\cdot 10^{-6}\) contradiction.
  4. True. \begin{align*} && (x^2-x^{-2})^2 &\geq 0 \\ \Leftrightarrow && x^4-2+x^{-4} &\geq0 \\ \Leftrightarrow && x^4+3+x^{-4} &\geq 5 \\ \end{align*}

View
1989 Paper 3 Q6
D: 1700.0 B: 1500.0
polar coordinates conics tangent-normal chord equation angle bisector trigonometric identities focus-directrix parametric

Show that, for a given constant \(\gamma\) \((\sin\gamma\neq0)\) and with suitable choice of the constants \(A\) and \(B\), the line with cartesian equation \(lx+my=1\) has polar equations \[ \frac{1}{r}=A\cos\theta+B\cos(\theta-\gamma). \] The distinct points \(P\) and \(Q\) on the conic with polar equations \[ \frac{a}{r}=1+e\cos\theta \] correspond to \(\theta=\gamma-\delta\) and \(\theta=\gamma+\delta\) respectively, and \(\cos\delta\neq0.\) Obtain the polar equation of the chord \(PQ.\) Hence, or otherwise, obtain the equation of the tangent at the point where \(\theta=\gamma.\) The tangents at \(L\) and \(M\) to a conic with focus \(S\) meet at \(T.\) Show that \(ST\) bisects the angle \(LSM\) and find the position of the intersection of \(ST\) and \(LM\) in terms of your chosen parameters for \(L\) and \(M.\)

Solution: \begin{align*} && \frac1{r} &= A \cos \theta + B \cos (\theta - \gamma) \\ &&&= A \cos \theta + B \cos \theta \cos \gamma + B \sin \theta \sin \gamma \\ &&&= (A+B \cos \gamma) \cos \theta + B \sin \gamma \sin \theta \\ \Longleftrightarrow && 1 &= (A+B \cos \gamma) x + B \sin \gamma y \end{align*} So if we choose \(B = \frac{m}{\sin \gamma}\) and \(A = l-m \cot \gamma\) we have the desired result. \begin{align*} && \frac{1 + e \cos (\gamma -\delta)}a &= A \cos (\gamma - \delta) + B \cos (\gamma - \delta - \gamma) \\ &&&= A \cos(\gamma-\delta) +B \cos \delta\\ && \frac{1 + e \cos (\gamma +\delta)}{a} &= A \cos (\gamma + \delta) + B \cos (\gamma + \delta - \gamma) \\ &&&= A \cos(\gamma + \delta) + B \cos \delta\\ \Rightarrow && \frac1{r} &= \frac{e}{a} \cos \theta + \frac{1}{a \cos \delta} \cos (\theta - \gamma) \\ \lim{\delta \to 0} &&\frac1{r} &= \frac{e}{a} \cos \theta+ \frac{1}{a} \cos (\theta - \gamma) \end{align*} Suppose we have have points \(L\) and \(M\) with \(\theta = \gamma_L, \gamma_M\) then our tangents are: \begin{align*} && \frac{a}{r} &= \cos \theta + \cos (\theta - \gamma_L) \\ && \frac{a}{r} &= \cos \theta + \cos (\theta - \gamma_M) \\ \Rightarrow && 0 &= \cos (\theta - \gamma_L) -\cos(\theta - \gamma_M) \\ &&&= - 2 \sin \frac{(\theta - \gamma_L)+(\theta - \gamma_M)}{2} \sin \frac{(\theta - \gamma_L)-(\theta - \gamma_M)}{2} \\ &&&= -2 \sin \left ( \theta - \frac{\gamma_L+\gamma_M}2 \right) \sin \left ( \frac{\gamma_M - \gamma_L}{2}\right) \\ \Rightarrow && \theta &= \frac{\gamma_L+\gamma_M}{2} \end{align*} Therefore clearly \(ST\) bisects \(LSM\). The line \(LM\) can be seen as the chord from the points \(\frac{\gamma_L+\gamma_M}{2} \pm \frac{\gamma_L-\gamma_M}{2}\), so the line is: \begin{align*} && \frac{a}{r} &= e \cos \theta + \frac{1}{\cos \left ( \frac{\gamma_L-\gamma_M}{2}\right)} \cos \left (\theta - \frac{\gamma_L+\gamma_M}{2} \right) \end{align*} and we want the point on the line where \(\theta =\frac{\gamma_L+\gamma_M}{2}\) so \begin{align*} && \frac{a}{r} &= e \cos \left ( \frac{\gamma_L+\gamma_M}{2} \right) + \frac{1}{\cos \left ( \frac{\gamma_L-\gamma_M}{2}\right)} \\ \Rightarrow && r &= \frac{a}{e \cos \left ( \frac{\gamma_L+\gamma_M}{2} \right) + \frac{1}{\cos \left ( \frac{\gamma_L-\gamma_M}{2}\right)}} \end{align*}

View