Problems

Solution: \begin{align*} && \cos(A \pm B) &= \cos A \cos B \mp \sin A \sin B \\ A = a, B = 3a&& \cos 4a + \cos 2a &= 2\cos 3a \cos a \\ \Rightarrow && \cos a \cos 3a &= \tfrac12(\cos 4a + \cos 2a) \\ \\ && \sin(A \pm B) &= \sin A \cos B \pm \cos A \sin B \\ && \sin 4a + \sin(- 2a) &= 2 \sin a \cos 3a \\ \Rightarrow && \sin a \cos 3a &= \tfrac12 (\sin 4a - \sin 2a) \end{align*}

1. \(\,\) \begin{align*} && 1 &= 4 \cos x \cos 2x \cos 3x \\ &&&= 2(\cos 4x +\cos 2x)\cos 2x \\ c = \cos 2x:&&&= 2(2c^2-1+c)c \\ \Rightarrow && 0 &= 4c^3+2c^2-2c-1 \\ &&&= (2c+1)(2c^2-1) \\ \Rightarrow && \cos 2x &= -\frac12 \\ \Rightarrow && x &= \frac{\pi}{3}, \frac{2\pi}{3} \\ && \cos 2x &= \pm \frac1{\sqrt2} \\ \Rightarrow && x&= \frac{\pi}{8}, \frac{3\pi}{8}, \frac{5\pi}{8}, \frac{7\pi}{8} \end{align*}
2. \(\,\) \begin{align*} && \tan x &= \tan 2x \tan 3x \tan 4x \\ \Rightarrow &&1 &= \frac{\cos x\sin 2x \sin 3x \sin 4x}{\sin x \cos 2x \cos 3x \cos 4x} \\ &&&= \frac{\sin 2x \sin 4x (\sin4 x + \sin 2x)}{\cos 2x \cos 4x (\sin 4x - \sin 2x)} \\ &&&= \frac{(\cos 2x - \cos 6x) (\sin4 x + \sin 2x)}{(\cos 6x + \cos 2x) (\sin 4x - \sin 2x)} \\ \Rightarrow && 0 &= 2\cos 6x \sin 4x - 2\cos 2x \sin 2 x\\ &&&= \sin 4 x (2 \cos 6x - 1) \\ \Rightarrow && \sin 4x &= 0 \\ \text{ or }&& \cos 6x &= \frac12 \end{align*} \(\sin 4x = 0 \Rightarrow x = 0, \frac{\pi}{4}, \frac{\pi}{2}, \frac{3\pi}{4}, \pi\) \(\cos 6x = \frac12 \Rightarrow x = \frac{\pi}{18}, \frac{5\pi}{18},\frac{7\pi}{18}, \frac{11\pi}{18}, \frac{13\pi}{18}, \frac{17\pi}{18}\). We should verify these work, since not all of them will, especially where \(\sin 4x = 0\), so our final answer is \(x = 0, \pi, \frac{\pi}{18}, \frac{5\pi}{18},\frac{7\pi}{18}, \frac{11\pi}{18}, \frac{13\pi}{18}, \frac{17\pi}{18}\)

Solution:

1. \(\,\) \begin{align*} && 0 &= \cos x + 3 \cos 2x + 3 \cos 3x + \cos 4 x \\ &&&= \cos x + \cos 4x + 3 \left (\cos 2x + \cos 3 x \right) \\ &&&= 2 \cos \frac{5x}{2} \cos \frac{3x}{2} + 6 \cos \frac{x}{2}\cos\frac{5x}{2} \\ &&&= 2 \cos \frac{5x}{2} \left (\cos \frac{3x}{2} + 3 \cos \frac{x}{2} \right)\\ &&&= 2 \cos \frac{5x}{2} \left ( \cos \frac{2x}{2}\cos \frac{x}{2} - \sin \frac{2x}{2} \sin \frac{x}{2}+3 \cos \frac{x}{2} \right) \\ &&&= 2 \cos \frac{5x}{2} \left ( \left (2\cos^2 \frac{x}{2} - 1 \right)\cos \frac{x}{2} - 2\sin \frac{x}{2} \cos \frac{x}{2} \sin \frac{x}{2}+3 \cos \frac{x}{2} \right) \\ &&&= 2 \cos \frac{5x}{2} \left ( 4\cos^3 \frac{x}{2} \right) \\ &&&= 8 \cos \frac{5x}{2} \cos^3 \frac{x}{2} \\ \Rightarrow && \frac{x}{2} &= \frac{\pi}{2}, \frac{3\pi}{2}, \cdots \\ && \frac{5x}{2} &= \frac{\pi}{2}, \frac{3\pi}{2}, \cdots \\ \Rightarrow && x &= \frac{\pi}{5}, \frac{3\pi}{5}, \pi, \frac{7\pi}{5}, \frac{9\pi}{5} \end{align*}
2. \(\,\) \begin{align*} && 1 &= \cos (x + y) + \cos(x-y) - \cos 2x \\ &&&= 2 \cos x \cos y - 2\cos^2 x + 1 \\ \Rightarrow && 0 &= \cos x (\cos y - \cos x) \\ \Rightarrow && 0 &=\cos x \left ( \cos y + \cos (\pi - x) \right) \\ &&&= 2\cos x \cos \frac{y+x-\pi}{2} \cos \frac{y-x+\pi}{2} \\ \Rightarrow && x &= \frac{\pi}{2} \\ && y+x - \pi&= \pi ,3\pi, \cdots \\ && y-x + \pi&=\pi, 3 \pi, \cdots \\ \Rightarrow && x &= \frac{\pi}{2} \\ && y+x &= 2\pi \Rightarrow x = y = \pi \\ && y&= x \end{align*} So the only solutions are \(x =y\) and \(x = \frac{\pi}{2}\)
3. \(\,\) \begin{align*} && \frac32 &= \cos x + \cos y - \cos (x+y) \\ &&&= 2 \cos \frac{x+y}{2} \cos \frac{x-y}{2} - 2 \cos^2 \frac{x+y}{2} + 1 \\ \Rightarrow && \frac14 &= \cos \frac{x+y}{2} \left ( \cos \frac{x-y}{2} - \cos \frac{x+y}{2} \right) \\ \Rightarrow && 0 &= \cos^2 \frac{x+y}{2} - \cos \frac{x-y}{2}\cos \frac{x+y}{2} + \frac14 \\ \Rightarrow && \cos \frac{x+y}{2} &= \frac{\cos \frac{x-y}{2} \pm \sqrt{\cos^2 \frac{x-y}{2}-1}}{2} \\ \Rightarrow && \cos \frac{x-y}{2} &= \pm 1\\ && \cos \frac{x+y}{2} &= \pm \frac12 \\ \Rightarrow && x-y &= -4\pi, 0, 4\pi, \cdots \\ \Rightarrow && x &= y \\ \Rightarrow && \cos x &= \frac12 \\ \Rightarrow && x &= \frac{\pi}{3} \\ \Rightarrow && (x, y) &= \left ( \frac{\pi}{3}, \frac{\pi}{3}\right) \end{align*}

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 \(\beta^2+\gamma^2 = \alpha^2\).

1. \(\,\) \begin{align*} && 2\sin(x-\tfrac14\pi) + \sqrt 3 \cos(x+\tfrac14\pi) &=\sin(x+\tfrac14\pi) \\ && 3 + 1 &= 4 \\ \Rightarrow && \left (\tan x + \frac{1+2}{\sqrt3} \right) \left ( \tan \frac{\pi}{4} + \frac{1-2}{\sqrt3}\right) &= 0\\ \Rightarrow && \tan x &= -\sqrt3 \\ \Rightarrow && x &= \tfrac23\pi, \tfrac53\pi \end{align*}
2. \(\,\) \begin{align*} && 2\sin(x-\frac16\pi) + \sqrt 3 \cos(x+\frac16\pi) &=\sin(x+\frac16\pi) \\ \Leftrightarrow && \left (\tan x + \frac{1+2}{\sqrt3} \right) \left ( \tan \frac{\pi}{3} + \frac{1-2}{\sqrt3}\right) &= 0\\ && x &\in [0, 2\pi) \end{align*}
3. \(\,\) \begin{align*} && 2\sin(x+\frac13\pi) + \sqrt 3 \cos(3x) = \sin(3x) \\ && A-B =x + \tfrac13\pi, A+B &= 3x \\ \Rightarrow && A = 2x + \tfrac\pi6, B &= x-\tfrac{\pi}{6} \\ \Rightarrow && \tan (2x+\tfrac\pi6)&=-\sqrt3 \\ && 2x + \tfrac{\pi}{6} &= \tfrac23\pi, \tfrac53\pi, \tfrac83 \pi, \tfrac{11}3\pi \\ && x &= \tfrac{\pi}{4}, \tfrac{3\pi}{4}, \tfrac{5\pi}{4}, \tfrac{7\pi}{4} \\ && \tan(-x-\tfrac{\pi}{6}) &= \frac1{\sqrt{3}} \\ \Rightarrow && x-\tfrac{\pi}{6} &= \ldots, \tfrac{\pi}{6}, \tfrac{7\pi}{6}, \ldots \\ \Rightarrow && x &= \tfrac{\pi}3, \tfrac{4\pi}{3} \\ \\ \Rightarrow && x &= \tfrac{\pi}{4}, \tfrac{3\pi}{4}, \tfrac{5\pi}{4}, \tfrac{7\pi}{4} , \tfrac{\pi}3, \tfrac{4\pi}{3} \end{align*}

Solution: \begin{align*} && \sin\left(r+\frac12\right)\theta - \sin\left(r-\frac12\right)\theta &= \sin r \theta \cos \tfrac12 \theta+\cos r \theta \sin \tfrac12 \theta- \left (\sin r \theta \cos \tfrac12 \theta-\cos r \theta \sin \tfrac12 \theta \right)\\ &&&= 2 \cos r\theta \sin \tfrac12 \theta \end{align*} \begin{align*} && S &= \cos a\theta + \cos (a + 1) \theta + \dots + \cos(b-2)\theta+\cos (b - 1 ) \theta \\ && 2\sin\tfrac12 \theta S &= \sum_{r=a}^{b-1} 2\sin\tfrac12 \theta \cos r \theta \\ &&&= \sum_{r=a}^{b-1} \left ( \sin\left(r+\frac12\right)\theta - \sin\left(r-\frac12\right)\theta \right) \\ &&&= \sin \left (b-\frac12 \right)\theta - \sin \left (a -\frac12 \right)\theta \\ \Rightarrow && \sin \left (b-\frac12 \right)\theta &= \sin \left (a -\frac12 \right)\theta \\ \end{align*} Case 1: \(A = B + 2n\pi\) \begin{align*} && \left (b-\frac12 \right)\theta &= \left (a -\frac12 \right)\theta + 2n\pi \\ \Rightarrow && (b-a) \theta &= 2n \pi \\ \Rightarrow && \theta &= \frac{2n\pi}{b-a} \end{align*} Case 2: \(A = (2n+1)\pi - B\) \begin{align*} && \left (b-\frac12 \right)\theta &= (2n+1)\pi -\left (a -\frac12 \right)\theta \\ \Rightarrow && (b+a-1) \theta &= (2n+1) \pi \\ \Rightarrow && \theta &= \frac{2n\pi}{b+a-1} \end{align*}

Solution: Let \(\tan \theta = t\) \begin{align*} \tan 3 \theta &\equiv \tan (2 \theta + \theta) \\ &\equiv \frac{\tan 2 \theta +\tan \theta}{1 - \tan 2 \theta \tan \theta} \\ &\equiv \frac{\frac{2t}{1-t^2}+t}{1-\frac{2t^2}{1-t^2}} \\ &\equiv \frac{2t+t-t^3}{1-t^2-2t^2} \\ &\equiv \frac{3t-t^3}{1-3t^3} \\ &\equiv \frac{3\tan \theta - \tan^3 \theta}{1 - 3 \tan^3 \theta} \end{align*} If \(\theta = \cos^{-1} (2/\sqrt{5})\), then \(\sin \theta = 1/\sqrt{5}\) and \(\tan \theta = 1/2\). Hence \begin{align*} \tan 3 \theta &= \frac{3 \cdot \frac12 - \frac18}{1 - \frac34} \\ &= \frac{11}{2} \end{align*}

1. Since \(\tan 3 y = 11/2\) has the solution \(y = \cos^{-1} (2/\sqrt{5})\) it will also have the solutions \(y = \cos^{-1}(2/\sqrt{5}) + \frac{\pi}{3}\) and \(y = \cos^{-1}(2/\sqrt{5})+\frac{2\pi}{3}\), therefore \begin{align*} && \cos^{-1} x &= \cos^{-1} (2/\sqrt{5})\\ \Rightarrow && x &= 2/\sqrt{5} \\ && \cos^{-1} x &= \cos^{-1} (2/\sqrt{5}) + \frac{\pi}{3}\\ \Rightarrow && x &= \frac{2}{\sqrt{5}} \frac{1}{2} - \frac{1}{\sqrt{5}} \frac{\sqrt{3}}{2} \\ &&&= \frac{2-\sqrt{3}}{2\sqrt{5}} \\ && \cos^{-1} x &= \cos^{-1} (2/\sqrt{5}) + \frac{2\pi}{3}\\ \Rightarrow && x &= \frac{2}{\sqrt{5}} \left (-\frac{1}{2} \right)- \frac{1}{\sqrt{5}} \frac{\sqrt{3}}{2} \\ &&&= \frac{-\sqrt{3}-2}{2\sqrt{5}} \\ \end{align*}
2. Since \(\cos \frac13 x = \frac{2}{\sqrt{5}}\) has the solution \(x = \tan^{-1} \frac{11}{2}\) it will also have the solutions \(x = \tan^{-1} \frac{11}{2} + 2n \pi\) and \(x = -\tan^{-1} \frac{11}{2} + 2n \pi\). \begin{align*} && \tan^{-1} y &= \tan^{-1} \frac{11}{2} \\ \Rightarrow && y &= \frac{11}{2} \\ && \tan^{-1} y &= \tan^{-1} \frac{11}{2} + 2n \pi \\ \Rightarrow && y &= \frac{\frac{11}{2} + 0}{1-0} \\ &&&= \frac{11}{2} \\ && \tan^{-1} y &= -\tan^{-1} \frac{11}{2} + 2n \pi \\ \Rightarrow && y &= \frac{-\frac{11}{2} + 0}{1-0} \\ &&&= -\frac{11}{2} \\ \end{align*} So our two solutions are \(y = \pm \frac{11}{2}\)