Problems

Solution: \begin{align*} \sum_{m=1}^n a_m(b_{m+1} -b_m) +\sum_{m=1}^n b_{m+1}(a_{m+1} -a_m) &= \sum_{m=1}^n \left (a_{m+1}b_{m+1}-a_mb_m \right) \\ &= a_{n+1}b_{n+1} - a_1b_1 \end{align*} And the result follows.

1. Let \(b_m = \sin m x \), \(a_m = \cosec \frac{x}{2}\), so \begin{align*} && \sum_{m=1}^n \cosec \frac{x}{2} \left (\sin (m+1)x - \sin mx \right) &= \sum_{m=1}^n \cosec \frac{x}{2} 2 \cos \left ( \frac{2m+1}{2}x \right) \sin \left ( \frac{(m+1)-m}{2}x \right) \\ &&&=2 \sum_{m=1}^n\cos \left ( (m + \tfrac12)x \right)\\ \\ \Rightarrow && \sum_{m=1}^n\cos \left ( (m + \tfrac12)x \right) &= \tfrac12 \cosec \tfrac{x}{2}\left ( \sin(n+1)x - \sin x \right) \end{align*}
2. \(\,\) \begin{align*} && b_{m+1}-b_m &= \sin m x \sin \tfrac12 x \\ &&&= \frac12 \left ( \cos (m-\tfrac12)x - \cos (m+\tfrac12)x \right)\\ \Rightarrow && b_m &= -\tfrac12 \cos (m - \tfrac12)x\\ && a_m &= m \\ \Rightarrow && \sum_{m=1}^n m \sin m x \sin \tfrac12 x &= (n+1) b_{n+1} - 1 \cdot b_1 - \sum_{m=1}^n b_{m+1} \cdot 1 \\ &&&= -(n+1) \tfrac12\cos(n+1-\tfrac12)x+\tfrac12\cos(\tfrac12x) + \tfrac12\sum_{m=1}^n \cos(m+\tfrac12)x \\ &&&= -(n+1) \tfrac12\cos(n+1-\tfrac12)x+\tfrac12\cos(\tfrac12x) + \tfrac14 \cosec \tfrac{x}{2}\left ( \sin(n+1)x - \sin x \right) \\ &&&= -(n+1) \tfrac12\cos(n+1-\tfrac12)x+ \tfrac14 \cosec \tfrac{x}{2}\sin(n+1)x \\ &&&= \tfrac12\cosec\tfrac{x}2 \left (\tfrac12 \sin (n+1)x-(n+1)\cos(n+\tfrac12)x\sin\tfrac12x \right) \\ &&&= \tfrac12\cosec\tfrac{x}2 \left (\tfrac12 \sin (n+1)x-(n+1)\tfrac12 \left ( \sin (n+1)x - \sin nx \right) \right) \\ &&&= \tfrac14 \cosec \tfrac{x}{2} \left ( -n \sin (n+1)x +(n+1) \sin n x \right) \end{align*} Therefore \(p = -\frac{n}4, q = \frac{n+1}{4}\)

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*}

1. \(\,\) \begin{align*} && 3 \cos 3 \theta &= \sin 3 \theta \left (\cot \theta - \cot (\tfrac13\pi - \theta) + \cot (\tfrac13\pi + \theta) \right) \\ \Rightarrow && 3 \cot 3\theta &= \cot \theta - \cot (\tfrac13\pi - \theta) + \cot (\tfrac13\pi + \theta) \\ \theta = \tfrac{\pi}{9}: && 3\cot \frac{\pi}{3} &= \cot \tfrac{\pi}{9} - \cot \tfrac{2}{9}\pi + \cot \tfrac49 \pi \\ \Rightarrow && \sqrt{3} &= \cot \tfrac{\pi}{9} - \cot \tfrac{2}{9}\pi + \cot \tfrac49 \pi \end{align*}
2. \(\,\) \begin{align*} \theta = \tfrac16\pi - \phi && \sin(\tfrac12\pi - 3\phi) &= 4\sin(\tfrac16\pi - \phi)\sin(\phi+\tfrac16\pi)\sin(\tfrac12\pi - \phi) \\ \Rightarrow && \cos 3\phi &= 4\cos(\phi - \tfrac13\pi)\cos(\tfrac13\pi - \phi)\cos\phi \\ \Rightarrow && \cot 3\theta &= \cot \theta\cot(\phi - \tfrac13\pi)\cot(\tfrac13\pi - \phi) \tag{dividing by (\(*\))} \\ \\ \frac{\d}{\d \theta}:&& -\csc^2 3\phi &= \cot3\phi \left (-\csc^2 \phi\tan \phi+\csc^2 (\tfrac13\pi - \phi) \tan (\tfrac13\pi - \phi) -\csc^2(\phi - \tfrac13\pi)\tan (\phi - \tfrac13\pi) \right) \\ \Rightarrow && \csc^23\phi\tan3\phi & = 2( \csc2\phi- \csc(\tfrac{2}{3}\pi - 2\phi)+\csc(\phi - \tfrac23\pi)) \\ \phi = \frac{1}{18}\pi: && 4\sqrt{3} &= 2(\csc \tfrac{1}{9}\pi - \csc \tfrac59\pi + \csc \tfrac79 \pi) \\ \end{align*} and the result follows.

Solution: \begin{align*} && I &= \int_{\frac{1}{6}\pi}^{\frac{1}{4}\pi} \frac 1{1-\cos2\theta} \;\d\theta \\ &&&= \int_{\frac{1}{6}\pi}^{\frac{1}{4}\pi} \frac 1{2\sin^2 \theta} \;\d\theta \\ &&&= \frac12\int_{\frac{1}{6}\pi}^{\frac{1}{4}\pi} \cosec^2 \theta \;\d\theta \\ &&&= \frac12\left [-\cot \theta \right]_{\frac{1}{6}\pi}^{\frac{1}{4}\pi} \\ &&&= \frac12 \left (\cot \frac{\pi}{6} - \cot \frac{\pi}{4} \right)\\ &&&= \frac{\sqrt{3} - 1}{2} \end{align*} \begin{align*} && J &= \int_{\sqrt3/2}^1 \frac 1 {1-\sqrt{1-x^2}} \, \d x \\ x = \sin 2 \theta, \d x = 2\cos 2\theta \d \theta &&&=\int_{\pi/6}^{\pi/4} \frac{2 \cos 2 \theta }{1-\cos 2 \theta} \d \theta \\ &&&=\int_{\pi/6}^{\pi/4} \frac{2 \cos 2 \theta -2+2}{1-\cos 2 \theta} \d \theta \\ &&&= -2\left (\frac{\pi}{4} - \frac{\pi}6 \right) + 2I \\ &&&= \sqrt{3}-1-\frac{\pi}{6} \end{align*} \begin{align*} && K &= \int_1^{2/\sqrt{3}} \frac{1}{y(y-\sqrt{y^2-1})} \d y \\ y = 1/x, \d x = -1/y^2 \d y &&&= \int_{x=1}^{x=\sqrt{3}/2} \frac{1}{1-\sqrt{1-x^2}} \d x\\ &&&= \sqrt{3}-1 -\frac{\pi}6 \end{align*}

Solution: \begin{align*} && \cot x - \tan x &= 2 \cot 2x \\ \Rightarrow && x\cot x - x\tan x &= 2x\cot 2x \\ \Rightarrow && 1 + \sum_{n=1}^{\infty} b_n x^n - \sum_{n=0}^{\infty}a_n x^{n+1} &= 1 + \sum_{n=1}^{\infty} b_n (2x)^n \\ \Rightarrow && \sum_{n=1}^{\infty}(1-2^n)b_nx^n &= \sum_{n=1}^{\infty} a_{n-1}x^n \\ \Rightarrow && a_{n-1} &= (1-2^n)b_n \quad \text{if }n \geq 1 \end{align*} \begin{align*} \cot x + \tan x &= 2 \cosec 2x \end{align*} So \begin{align*} && \cot x + \tan x &= 2 \cosec 2x \\ \Rightarrow && x \cot x + x\tan x &= 2x \cosec 2x \\ \Rightarrow && 1 + \sum_{n=1}^{\infty} b_n x^n + \sum_{n=0}^{\infty} a_n x^{n+1} &= 1+\sum_{n=1}^\infty c_n (2x)^n \\ \Rightarrow && \sum_{n=1}^{\infty} \frac{1}{1-2^n}a_{n-1} +\sum_{n=1}^{\infty}a_{n-1}x^n &= \sum_{n=1}^{\infty} 2^nc_n x^n \\ \Rightarrow && c_n &= \frac{1}{2^n} \left ( 1 + \frac{1}{1-2^n} \right)a_{n-1} \\ &&&= \frac1{2^n} \frac{2^n-2}{2^n-1} a_{n-1}\\ &&&= \frac1{2^{n-1}}\frac{2^{n-1}-1}{2^n-1} a_{n-1} \end{align*}