Problems
Filters
1 problem found
2005 Paper 3 Q1
Show that \(\sin A = \cos B\) if and only if \(A = (4n+1)\frac{\pi}{2}\pm B\) for some integer \(n\). Show also that \(\big\vert\sin x \pm \cos x \big\vert \le \sqrt{2}\) for all values of \(x\) and deduce that there are no solutions to the equation \(\sin\left( \sin x \right) = \cos \left( \cos x \right)\). Sketch, on the same axes, the graphs of \(y= \sin \left( \sin x \right)\) and \(y = \cos \left( \cos x \right)\). Sketch, not on the previous axes, the graph of \(y= \sin \left(2 \sin x \right)\).
Solution: \begin{align*} && \sin A &= \cos B \\ \Leftrightarrow && 0 &= \sin A - \cos B \\ &&&= \sin A - \sin ( \frac{\pi}{2} - B) \\ &&&= 2 \sin \left ( \frac{A + B - \frac{\pi}{2}}{2} \right) \cos \left (\frac{A - B + \frac\pi2}{2} \right) \\ \Leftrightarrow && n \pi &= \frac{A+B - \frac{\pi}{2}}{2}, n\pi + \frac{\pi}{2} = \frac{A-B+\frac{\pi}{2}}{2} \\ \Leftrightarrow && A \pm B &= 2n\pi + \frac{\pi}{2} \\ &&&= (4n+1) \frac{\pi}{2} \end{align*} \begin{align*} |\sin x \pm \cos x| &= | \sqrt{2} \sin(x \pm \frac{\pi}{4} )| \\ & \leq \sqrt{2} \end{align*} Therefore if \(\sin(\sin x) = \cos (\cos x)\) we must have that \(|\sin x \pm \cos x| = |(4n+1) \frac{\pi}{2}| \geq \frac{\pi}{2} > 1.5 > \sqrt{2}\) contradiction.
