# Value of $\arg(\sin \theta +i\cos \theta)$

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 $$\pi$$