Problems

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Solution:

1. If we can show that \(0 < a - \sqrt{a^2-b} < 1\) then we will be done, since raising a number in \([0,1)\) to a positive integer power will always remain in the same interval. Clearly \(\sqrt{a^2-b} < \sqrt{a^2} = a\) so we have \(a-\sqrt{a^2-b} > 0\) We also have that \(\sqrt{a^2-b} > \sqrt{a^2-(2a-1)} = (a-1)\). Therefore \(a - \sqrt{a^2-b} < a - (a-1) = 1\) as required.
2. If we can show that \(\l a + \sqrt{a^2-b} \r^n + \l a - \sqrt{a^2-b} \r^n = N -r + s\) is an integer we will be done, since the only integer value \(-r+s\) can be is \(0\). This is easy to see, since \begin{align*} \l a + \sqrt{a^2-b} \r^n + \l a - \sqrt{a^2-b} \r^n &= \sum_{k=0}^n \binom{n}{k}a^{n-k}(\sqrt{a^2-b})^k +\sum_{k=0}^n \binom{n}{k}a^{n-k}(-\sqrt{a^2-b})^k \\ &= \sum_{k=0}^n \binom{n}{k}a^{n-k}\l (\sqrt{a^2-b})^k +(-\sqrt{a^2-b})^k \r \\ \end{align*} But every term where \(k\) is odd in this sum is \(0\) (since they cancel) and ever term where \(k\) is even in this sum is an integer. Therefore the sum is an integer and we're done.
3. \begin{align*} -r^2+rN &= -r(r-N) \\ &= s(r-N) \\ &=- \l a - \sqrt{a^2-b} \r^n \l a + \sqrt{a^2-b} \r^n \\ &= -\l \l a - \sqrt{a^2-b} \r \l a + \sqrt{a^2-b} \r\r^n \\ &= - \l a^2 - a^2+b\r^n \\ &= b^n \end{align*} Therefore \(r^2-rN + b^n = 0\)

Looking at \(\left(8+3\sqrt{7}\right)^{n}\) we have \(a = 8, b = 1\) (since \(8^2 - 1 = 9 \cdot 7\). So we can apply the result of the previous question to see that: \(\left(8+3\sqrt{7}\right)^{n}\) differs from an integer by \(\left(8-3\sqrt{7}\right)^{n}\). \begin{align*} 8-3\sqrt{7} &= \frac{1}{8+3\sqrt{7}} \\ &\approx \frac{1}{8 + 8} \\ &\approx 2^{-4} \end{align*} Therefore it differs by approximation \((2^{-4})^n = 2^{-4n}\)

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Solution: The point \(ze^{i\theta}\) is obtained by rotating the point \(z\) about \(0\) by an angle \(\theta\) anticlockwise. The complex numbers \(\alpha, \beta\) and \(\gamma\) will form an equilateral triangle iff the angles between each side are \(\frac{\pi}{3}\), ie \begin{align*} \begin{cases}{\gamma - \beta} &= e^{i \frac{\pi}{3}}({\beta - \alpha}) \\ {\alpha- \gamma} &= e^{i \frac{\pi}{3}}({\gamma- \beta}) \\ {\beta- \alpha} &= e^{i \frac{\pi}{3}}({\alpha- \gamma})\end{cases} \end{align*} We don't need all these equations, since the first two are equivalent to the third. Combining the first two equations, we have \begin{align*} && \frac{\gamma - \beta}{\beta-\alpha} &= \frac{\alpha-\gamma}{\gamma - \beta} \\ \Leftrightarrow && (\gamma - \beta)^2 &= (\alpha-\gamma)(\beta-\alpha) \\ \Leftrightarrow && \gamma^2 +\beta^2 - 2\gamma \beta &= \alpha\beta-\alpha^2-\gamma\beta+\gamma\alpha \\ \Leftrightarrow && \alpha^{2}+\beta^{2}+\gamma^{2}-\beta\gamma-\gamma\alpha-\alpha\beta&=0 \end{align*} as required. If the roots of \(z^{3}+az^{2}+bz+c=0\) are \(\alpha, \beta, \gamma\) then \(\alpha+\beta+\gamma = -a\) and \(\beta\gamma+\gamma\alpha+\alpha\beta = b\). We also have that \(a^2 - 2b = \alpha^2+\beta^2+\gamma^2\). Therefore there roots will lie at the vertices of an equilateral triangle iff \(a^2-3b = 0\)

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Solution:

1. If \(f(x) = \alpha x\) then \(f^{-1}(x) = \frac{1}{\alpha}x\). \begin{align*} && \frac{\d^{n-1}}{\d y^{n-1}} [y - \alpha y]^n &= \frac{\d^{n-1}}{\d y^{n-1}} [(1 - \alpha)^n y^n] \\ && &= (1-\alpha)^n n! y \\ \Rightarrow && y + \sum_{n=1}^{\infty} \frac{1}{n!}\frac{\d^{n-1}}{\d y^{n-1}} [y - \alpha y]^n &= y +\sum_{n=1}^{\infty} (1-\alpha)^ny \\ &&&= y + y\l \frac{1}{1-(1-\alpha)}-1 \r \\ &&&= \frac{1}{\alpha}y \end{align*} Where we can sum the geometric progression if \(|1-\alpha| < 1 \Leftrightarrow 0 < \alpha < 2\)
2. Suppose that \(f(x) = x-\frac14x^3\). We would like to find \(f^{-1}(\frac12)\). \begin{align*} && \frac{\d^{n-1}}{\d y^{n-1}} [y - (y+\frac14 y^3)]^n &= \frac{\d^{n-1}}{\d y^{n-1}} [\frac1{4^n} y^{3n}] \\ && &= \frac{1}{4^n} \frac{(3n)!}{(2n+1)!} y^{2n+1} \\ \Rightarrow && f^{-1}(\frac12) &= \frac12 + \sum_{n=1}^{\infty} \frac{1}{4^n} \frac{(3n)!}{n!(2n+1)!} \frac{1}{2^{2n+1}} \\ &&&= \frac12 + \sum_{n=1}^{\infty} \frac{(3n)!}{n!(2n+1)!} \frac{1}{2^{4n+1}} \\ \end{align*} Since when \(n = 0\) \(\frac{0!}{0!1!} \frac{1}{2^{0+1}} = \frac12\) we can include the wayward \(\frac12\) in our infinite sum and so we have the required result.
3. Consider \(f(x) = x - e^{\lambda x}\) we are interested in \(f^{-1}(0)\). \begin{align*} && \frac{\d^{n-1}}{\d y^{n-1}} [y - (y-e^{\lambda y})]^n &= \frac{\d^{n-1}}{\d y^{n-1}} [e^{n\lambda y}] \\ &&&= n^{n-1} \lambda^{n-1}e^{n \lambda y} \\ \Rightarrow && f^{-1}(0) &= \sum_{n=1}^\infty \frac{1}{n!} n^{n-1} \lambda^{n-1} \end{align*} We don't care about convergence, but it's worth noting this has a radius of convergence of \(\frac{1}{e}\) (ie this series is valid if \(|\lambda| < \frac1e\)).

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Solution: \begin{align*} \int_{-\frac{1}{2}\pi}^{\frac{1}{2}\pi}\frac{\cos^{2}\theta}{1-\sin\theta\sin2\alpha}\,\mathrm{d}\theta &= \int_{u = \frac12 \pi}^{u = -\frac12 \pi} \frac{\cos^2 (-u)}{1-\sin(-u) \sin 2 \alpha} -\d u \tag{\(u = -\theta\)} \\ &= \int_{\frac12 \pi}^{-\frac12 \pi} \frac{\cos^2 u}{1+\sin u \sin 2 \alpha} -\d u \\ &= \int_{-\frac12 \pi}^{\frac12 \pi} \frac{\cos^2 u}{1+\sin u \sin 2 \alpha} \d u \\ &= \int_{-\frac12 \pi}^{\frac12 \pi} \frac{\cos^2 \theta}{1+\sin \theta \sin 2 \alpha} \d \theta \\ \end{align*} Since \(\displaystyle \frac{1}{(1-a^2u^2)} = \frac12 \l \frac{1}{1+au} + \frac1{1-au} \r\) \begin{align*} \int_{-\frac12 \pi}^{\frac12 \pi} \frac{\cos^2 \theta}{1+\sin \theta \sin 2 \alpha} \d \theta &= \int_{-\frac12 \pi}^{\frac12 \pi} \frac{\cos^2 \theta}{1-\sin ^2\theta \sin^2 2 \alpha} \d \theta \\ &= \int_{-\frac12 \pi}^{\frac12 \pi} \frac{1-\sin^2 \theta}{1-\sin ^2\theta \sin^2 2 \alpha} \d \theta \\ &= \int_{-\frac12 \pi}^{\frac12 \pi} \frac{(1-\sin ^2\theta \sin^2 2 \alpha) \frac{1}{\sin^2 2\alpha} + 1 - \cosec^2 2\alpha}{1-\sin ^2\theta \sin^2 2 \alpha} \d \theta \\ &= \frac{\pi}{\sin^2 2\alpha} -\cot^2 2\alpha \int_{-\frac{\pi}2}^{\frac{\pi}2} \frac{1}{1 - \sin^2 \theta \sin^2 2 \alpha} \d \theta \\ &= \frac{\pi}{\sin^2 2\alpha} -\cot^2 2\alpha \int_{-\frac{\pi}2}^{\frac{\pi}2} \frac{1}{1 - \sin^2 \theta (1-\cos^2 2 \alpha)} \d \theta \\ &= \frac{\pi}{\sin^2 2\alpha} -\cot^2 2\alpha \int_{-\frac{\pi}2}^{\frac{\pi}2} \frac{1}{\cos^2 \theta +\sin^2 \theta \cos^2 2 \alpha} \d \theta \\ &= \frac{\pi}{\sin^2 2\alpha} -\cot^2 2\alpha \int_{-\frac{\pi}2}^{\frac{\pi}2} \frac{\sec^2 \theta}{1 +\tan^2 \theta \cos^2 2 \alpha} \d \theta \\ \end{align*} Finally, using the substitution \(u =|\cos 2 \alpha | \tan \theta, \d u = |\cos 2 \alpha |\sec^2 \theta \d \theta\) \begin{align*} \int_{-\frac{\pi}2}^{\frac{\pi}2} \frac{\sec^2 \theta}{1 +\tan^2 \theta \cos^2 2 \alpha} \d \theta &= |\sec 2\alpha|\int_{u = -\infty}^{u = \infty} \frac{1}{1 + u^2} \d u \\ &= |\sec 2 \alpha|\pi \end{align*} and so \begin{align*} I &= \frac{\pi}{\sin^2 2\alpha} -\cot^2 2\alpha |\sec 2 \alpha|\pi \\ &= \frac{\pi}{\sin^2 2\alpha} \l 1-\cos 2\alpha \r \\ &= \frac{\pi}{4\sin^2 \alpha\cos^2 \alpha} \l 2 \sin^2 \alpha \r \\ &= \frac{\pi}{2 \cos^2 \alpha} = \frac{\pi}{2} \sec^2 \alpha \end{align*} When \(\alpha\) small enough that the modulus doesn't flip the sign. When if \(\frac{1}{4}\pi<\alpha<\frac{1}{2}\pi\) we have: \begin{align*} I &= \frac{\pi}{\sin^2 2\alpha} -\cot^2 2\alpha |\sec 2 \alpha|\pi \\ &= \frac{\pi}{\sin^2 2\alpha} \l 1+\cos 2\alpha \r \\ &= \frac{\pi}{4\sin^2 \alpha\cos^2 \alpha} \l 2 \cos^2 \alpha \r \\ &= \frac{\pi}{2 \sin^2 \alpha} = \frac{\pi}{2} \cosec^2 \alpha \end{align*}

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Solution:

We can approximate the integral by \(N\) trapeziums, each with height \(x_{i+1}-x_{i} = \frac{x_N-x_0}{N} = \frac{h}{N}\). The will have area \(\frac{(f(x_i)+f(x_{i+1}))h}{2}\) and summing all these areas we will get: \[\frac12 h \l f(x_0) + f(x_1) + f(x_1)+f(x_2) + \cdots + f(x_{N-1})+f(x_N) \r = \frac12 h \l f(x_0) +2 f(x_1) + + \cdots +2f(x_{N-1})+f(x_N) \r\] But this is approximately the integral \(\displaystyle \int_{x_0}^{x_N} f(x) \d x\) \begin{align*} && \int_1^n \ln x \d x &= [x \ln x]_1^n - \int_1^n x \cdot \frac{1}{x} \d x \\ &&&= n \ln n - n+1 \\ &&&\approx \frac12 \l \ln 1 + 2\sum_{k=2}^{n-1} \ln k + \ln n \r \\ &&&= \ln (n!) - \frac12 \ln n \\ \Rightarrow && \ln (n!) &\approx n \ln n + \frac12 \ln n - n + 1 \\ \Rightarrow && n! &\approx \exp(n \ln n + \frac12 \ln n - n + 1) \\ &&&=n^{n+\frac12}e^{1-n} \end{align*} Since \(\ln x\) is a concave function, we should expect all the trapeziums to all lie under the curve, therefore this is always an underestimate for the integral, ie \(n! < g(n)\) \begin{align*} && \int_1^n \ln x \d x &= n \ln n - n+1 \\ &&&\approx \frac12 k^{-1} \l \ln 1 + 2\sum_{r=1}^{k(n-1)-1} \ln \l 1+\frac{r}{k} \r + \ln n \r \\ &&&=\frac{1}{2k} \l 2\sum_{r=1}^{k(n-1)-1} \l \ln(k+r) - \ln k)\r + \ln n\r \\ &&&=\frac1{k} \l \ln ((k+k(n-1)-1)!) - \ln(k!) - (k(n-1)-1) \ln k+\frac12\ln n \r \\ &&&=\frac1{k} \l \ln ((kn-1)!) - \ln(k!) - (k(n-1)-1) \ln k+\frac12 \ln n \r \\ &&&=\frac1{k} \l \ln ((kn)! ) -\ln k -\ln n - \ln(k!) - (k(n-1)-1) \ln k+\frac12\ln n \r \\ &&&= \frac1{k} \l \ln ((kn)! ) - \ln(k!) - (k(n-1)) \ln k - \frac12 \ln n\r \\ \Rightarrow && \ln ((kn)!) &\approx kn \ln n - kn + k + \ln(k!) + (k(n-1)) \ln k + \frac12 \ln n\\ \Rightarrow && (kn)! &\approx n^{kn+\frac12}e^{-k(n-1)}k!k^{k(n-1)} \\ &&&= n^{kn+\frac12} k! \l \frac{e}{k} \r^{k(1-n)} \end{align*} I would expect this approximation to be a better approximation for \((kn)!\) since it is created using a finer mesh.

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Solution:

1. \(\mathbf{n} \cdot \mathbf{r} = 0\) is the equation of a plane with normal \(\mathbf{n}\). \(\mathbf{n} \cdot (\mathbf{r}-\mathbf{a}) = 0\) is the equation of a plane through \(\mathbf{a}\) with normal \(\mathbf{n}\). Our expression is: \begin{align*} && \left(\mathbf{a-b}\right) \cdot \mathbf{r}&=\frac{1}{2}(\left|\mathbf{a}\right|^{2}-\left|\mathbf{b}\right|^{2}) \\ &&&=\frac{1}{2}(\mathbf{a}-\mathbf{b})\cdot(\mathbf{a}+\mathbf{b}) \\ \Leftrightarrow && \left(\mathbf{a-b}\right) \cdot \left ( \mathbf{r} - \frac12 (\mathbf{a}+\mathbf{b}) \right) &= 0 \end{align*} So this is a plane through \(\frac12 (\mathbf{a}+\mathbf{b})\) perpendicular to \(\mathbf{a}-\mathbf{b}\). ie the plane halfway between \(\mathbf{a}\) and \(\mathbf{b}\) perpendicular to the line between them.
2. \begin{align*} && 0 &= \left(\mathbf{a-r}\right)\cdot\left(\mathbf{b-r}\right) \\ &&&= \mathbf{a} \cdot \mathbf{b} - \mathbf{r} \cdot (\mathbf{a}+\mathbf{b}) + \mathbf{r}\cdot\mathbf{r} \\ &&&= \left ( \mathbf{r}- \frac12(\mathbf{a}+\mathbf{b}) \right) \cdot \left ( \mathbf{r}- \frac12(\mathbf{a}+\mathbf{b}) \right) - \frac14 \left (\mathbf{a}\cdot\mathbf{a}+2\mathbf{a}\cdot\mathbf{b} + \mathbf{b}\cdot\mathbf{b} \right) +\mathbf{a}\cdot\mathbf{b} \\ &&&= \left | \mathbf{r} - \frac12 \left (\mathbf{a}+\mathbf{b} \right) \right|^2 - \left |\frac12 \left ( \mathbf{a} - \mathbf{b}\right) \right|^2 \end{align*} Therefore this is a sphere, centre \(\frac12 \left (\mathbf{a}+\mathbf{b} \right)\) radius \(\left |\frac12 \left ( \mathbf{a} - \mathbf{b}\right) \right|\)
3. This is a sphere centre \(\mathbf{a}\) radius \(\frac1{\sqrt{2}} \left|\mathbf{a-b}\right|\)
4. This is a sphere centre \(\mathbf{b}\) radius \(\frac1{\sqrt{2}} \left|\mathbf{a-b}\right|\)

Suppose the first two cases are true, then by symmetry it suffices to show that we can prove either of the second cases are true. (Since everything is symmetric in \(\mathbf{a}\) and \(\mathbf{b}\)). It's useful to note that \(\mathbf{r}\cdot \mathbf{r} = \mathbf{r}\cdot \mathbf{b} + \mathbf{r}\cdot \mathbf{a} -\mathbf{a}\cdot\mathbf{b}\) from the second condition. \begin{align*} \left|\mathbf{r-a}\right|^{2} &= \mathbf{r} \cdot \mathbf{r}-2\mathbf{a}\cdot \mathbf{r} + \mathbf{a}\cdot \mathbf{a} \\ &= \mathbf{r}\cdot \mathbf{b} + \mathbf{r}\cdot \mathbf{a} -\mathbf{a}\cdot\mathbf{b} - 2\mathbf{a}\cdot \mathbf{r} + \mathbf{a}\cdot \mathbf{a} \\ &= \mathbf{r} \cdot ( \mathbf{b} - \mathbf{a}) + \mathbf{a} \cdot (\mathbf{a}-\mathbf{b}) \\ &= -\frac{1}{2}(\left|\mathbf{a}\right|^{2}-\left|\mathbf{b}\right|^{2}) + |\mathbf{a}|^2- \mathbf{a}\cdot\mathbf{b} \\ &= \frac{1}{2} |\mathbf{a}-\mathbf{b}|^2 \end{align*} as required. To show the other direction, consider Geometrically, these cases are equivalent, because together they both describe a circle of radius \(\left |\frac12 \left ( \mathbf{a} - \mathbf{b}\right) \right|\) in the plane halfway between \(\mathbf{a}\) and \(\mathbf{b}\)

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Solution: We use the following properties: \((\mathbf{AB})^T = \mathbf{B}^T\mathbf{A}^T\), \((\mathbf{AB})^{-1} = \mathbf{B}^{-1}\mathbf{A}^{-1}\), and \((\mathbf{A}^T)^{-1} = (\mathbf{A}^{-1})^T\) \begin{align*} &&\mathbf{I} &= \mathbf{B}^{\mathrm{T}}\mathbf{B} \\ \Leftrightarrow && {\mathbf{B}^{T}}^{-1} &= \mathbf{B} \\ \Leftrightarrow && (\mathbf{I+A})(\mathbf{I-A})^{-1} &= {((\mathbf{I+A})(\mathbf{I-A})^{-1})^{T}}^{-1} \\ &&&= {((\mathbf{I+A})(\mathbf{I-A})^{-1})^{-1}}^{T} \\ &&&= ((\mathbf{I-A})(\mathbf{I+A})^{-1})^{T} \\ &&&= {(\mathbf{I+A})^{-1}}^T(\mathbf{I-A})^T \\ &&&= {(\mathbf{I+A}^T)}^{-1}(\mathbf{I-A}^T) \\ \Leftrightarrow && (\mathbf{I+A}^T)(\mathbf{I+A}) &= (\mathbf{I-A}^T)(\mathbf{I-A}) \\ \Leftrightarrow && \mathbf{I}+\mathbf{A}+\mathbf{A}^T+\mathbf{A}^T\mathbf{A} &= \mathbf{I}-\mathbf{A}-\mathbf{A}^T+\mathbf{A}^T\mathbf{A} \\ \Leftrightarrow && 2( \mathbf{A}^T+\mathbf{A}) &= \mathbf{O} \\ \Leftrightarrow && \mathbf{A}+\mathbf{A}^T &= \mathbf{O} \end{align*}

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Solution:

1. (Closure) This operation is clearly closed by construction
2. (Associative) \(a_j \circ (a_k \circ a_l) = a_j \circ a_{\max(k,l)} = a_{\max(j,k,l)} = a_{\max(j,k)} \circ a_l = (a_j \circ a_k) \circ a_l\), so it is associative
3. (Identity) \(a_1 \circ a_k = a_{\max(1,k)} = a_k = a_{\max(k,1)} = a_k \circ a_1\) so \(a_1\) is an identity.
4. (Inverses) There is no inverse, since \(a_N \circ a_k = a_N\) for all \(k\), and hence \(a_N\) can have no inverse.

1. (Closure) \(n = |j-k|+1 \geq 1\) so we need to show that \(n \leq N\) to ensure closure. This is true since the largest \(j-k\) can be is if \(j = N\) and \(k = 1\), and this also satisfies \(|j-k| + 1 \leq N\). Hence the operation is closed.
2. (Associative) \(a_j * (a_k * a_l) = a_j * (a_{|k-l|+1}) = a_{|j-|k-l|-1|+1}\). \((a_j * a_k) * a_l = a_{|j-k|+1}*a_l = a_{|l-|j-k|-1|+1}\). \(a_2 * (a_2 * a_3) = a_2 * a_2 = a_1\). \((a_2 *a_2)*a_3 = a_1 * a_3 = a_3 \neq a_2\) therefore this isn't associative for any \(N > 2\)
3. (Identity) \(a_1\) is an identity, since \(a_1 * a_k = a_{|k-1|+1} = a_{k-1+1} = a_k\).
4. (Inverse) Every element is self-inverse since \(a_k * a_k = a_{|k-k|+1} = a_1\)

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Solution:

It is important to define clear coordinate axes, so let the \(x\)-axis point up the line of greatest slope of the ring. The \(z\)-axis perpendicular to the ring, and the \(y\)-axis perpendicular to both of these. Our method is going to be to take moments about \(O\) to avoid worrying about the force at the pivot. There are \(3\) forces we need to worry about:

• The mass of the rod
• The reaction where it meets the ring
• The friction at the ring

In our coordinate frame, the reaction will act in the \(z\)-direction, \(\displaystyle \begin{pmatrix} 0 \\ 0 \\ R \end{pmatrix}\), the friction force will act in the \(x-y\) plane: \(\displaystyle \begin{pmatrix} \mu R \sin \theta \\ -\mu R \cos \theta \\ 0 \end{pmatrix}\). We don't know the mass, but we know it will be acting "vertically", so \(\cos \alpha\) of it will act in the \(z\)-axis and \(\sin \alpha\) will act in the \(y\)-axis, ie it will act parallel to \(\displaystyle \begin{pmatrix} \sin \alpha \\ 0 \\ \cos \alpha \end{pmatrix}\). When taking moments, we need to consider \(\mathbf{r}\) the direction of the rod. This will be \(\displaystyle \begin{pmatrix} \cos \theta \\ \sin \theta \\ 0 \end{pmatrix}\). The moment of the weight will all be parallel to \(\mathbf{r} \times \begin{pmatrix} \sin \alpha \\ 0 \\ \cos \alpha \end{pmatrix}\). Similarly the moments of the contact forces will be \(\mathbf{r} \times \begin{pmatrix} \mu R \sin \theta \\ -\mu R \cos \theta \\ R \end{pmatrix}\). Since these moments sum to \(\mathbf{0}\) as we are in equilibrium, these vectors must be parallel. Therefore it is sufficient to check the vector triple product, \begin{align*} && 0 &= \begin{pmatrix} \cos \theta \\ \sin \theta \\ 0 \end{pmatrix} \cdot \left ( \begin{pmatrix} \sin \alpha \\ 0 \\ \cos \alpha \end{pmatrix} \times \begin{pmatrix} \mu \sin \theta \\ -\mu \cos \theta \\ 1 \end{pmatrix} \right ) \\ &&&= \cos \theta (\mu \cos \theta \cos \alpha)-\sin \theta (\sin \alpha - \mu \sin \theta \cos \alpha) \\ &&&= \mu((\sin^2 \theta+\cos^2 \theta) \cos \alpha) -\sin \theta \sin \alpha \\ \Rightarrow && \mu &= \tan \alpha \sin \theta \end{align*}