Problems

Filters
Clear Filters

1 problem found

2018 Paper 2 Q4
D: 1600.0 B: 1484.0
trigonometry sum-to-product formula cosine solving trigonometric equations product-to-sum formula simultaneous equations optimisation

In this question, you may use the following identity without proof: \[ \cos A + \cos B = 2\cos\tfrac12(A+B) \, \cos \tfrac12(A-B) \;. \]

  1. Given that \(0\le x \le 2\pi\), find all the values of \(x\) that satisfy the equation \[ \cos x + 3\cos 2x + 3\cos 3 x + \cos 4x= 0 \,. \]
  2. Given that \(0\le x \le \pi\) and \(0\le y \le \pi\) and that \[ \cos (x+y) + \cos (x-y) -\cos2x = 1 \,, \] show that either \(x=y\) or \(x\) takes one specific value which you should find.
  3. Given that \(0\le x \le \pi\) and \(0\le y \le \pi\,\), find the values of \(x\) and \(y\) that satisfy the equation \[ \cos x + \cos y -\cos (x+y) = \tfrac32 \,. \]

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*}

View