Problem source

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 source

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.