If $\displaystyle \frac{3+i\sin \theta}{4-i\cos \theta}$ is purely real number , where $\theta \in [0,2\pi].$ Then what is $\arg(\sin \theta +i\cos \theta)$?

What I tried:

\begin{align*} \frac{3+i\sin \theta}{4-i\cos \theta} & =\frac{(3+i\sin \theta)(4+i\cos \theta)}{(4-i\cos \theta)(4+i\cos \theta)}\\ &=\frac{12-(\sin \theta\cos \theta)+i(4\sin \theta+3\cos \theta)}{16+\cos^2 \theta}\in \mathbb{R} \end{align*}

means $(4\sin \theta+3\cos \theta)=0$, namely $\displaystyle \tan \theta = -3/4$. So either $\theta\in (\pi/2,\pi)$ or $\theta\in(3\pi/2,2\pi)$.

Now $\arg(\sin \theta+i\cos\theta)=\arctan\left(\frac{\cos \theta}{\sin \theta}\right)=\arctan(\cot\theta)=-4/3$, but the answer given as $\displaystyle \pi-\tan^{-1}(4/3)$.

How do I solve this? Help me please.


So your work is good until the third from last line. You forgot the arctan. So your solution should be $\arctan(\frac{-4}{3})$ now remember arctan is an odd function so $\arctan(\frac{-4}{3})=-\arctan(\frac{4}{3})$ which will be a negative number and you want $\theta \in [0,2 \pi]$ So we add $\pi$ to get what we want. We can do this because tangent is periodic of period $\pi$

Hope this helps.


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