1 problem found
Write down a value of \(\theta\,\) in the interval \(\frac{1}{4}\pi< \theta <\frac{1}{2}\pi\) that satisfies the equation \[ 4\cos\theta+ 2\sqrt3\, \sin\theta = 5 \;. \] Hence, or otherwise, show that \[ \pi=3\arccos(5/\sqrt{28}) + 3\arctan(\sqrt3/2)\;. \] Show that \[ \pi=4\arcsin(7\sqrt2/10) - 4\arctan(3/4)\;. \]
Solution: If \(\theta = \frac{\pi}{3}\) then \(\cos \theta = \frac12, \sin \theta = \frac{\sqrt{3}}{2}\) and clearly the equation is satisfied. We can also solve this equation using the harmonic formulae, namely: \begin{align*} && 5 &= 4 \cos \theta + 2\sqrt{3} \sin \theta \\ &&&= \sqrt{4^2+2^2 \cdot 3} \cos \left (\theta -\tan^{-1} \left (\frac{2\sqrt{3}}{4}\right) \right) \\ \Rightarrow && \frac{5}{\sqrt{28}} &= \cos \left ( \frac{\pi}{3} - \tan^{-1} \left (\frac{\sqrt{3}}{2}\right) \right) \\ \Rightarrow && \frac{\pi}{3} &= \arccos\left( \frac{5}{\sqrt{28}}\right) + \arctan \left (\frac{\sqrt{3}}{2}\right) \end{align*} From which the result follows. Similarly, notice that \(3 \cos \theta + 4 \sin \theta = \frac{7}{\sqrt{2}}\) is clearly solved by \(\frac{\pi}{4}\), but also writing it in harmonic form, we have \begin{align*} &&\frac{7}{\sqrt{2}} &= 5 \sin \left (\theta + \tan^{-1} \left ( \frac{3}{4} \right) \right) \\ \Rightarrow && \frac{7\sqrt{2}}{10} &= \sin \left ( \frac{\pi}{4} + \tan^{-1} \left ( \frac{3}{4} \right) \right) \\ \Rightarrow && \frac{\pi}{4} &= \arcsin \left ( \frac{7\sqrt{2}}{10} \right) - \arctan \left ( \frac{3}{4} \right) \end{align*} as required.