This is the question in our assignment, but I just want to know if this question really makes sense.
Wouldn't be the unit inside the $\sin$ be $\frac{1}{°}$ so it doesn't make sense? Should I convert it to radians and just continue?
Suppose that $90°<\theta<180°$. the sign of $\sin\frac{2}{\theta}$
trigonometry
Related Solutions
From Gerry Myerson's comment, we have \begin{align*} \frac{18\pi}{8} &= \frac{9\pi}{4} \\ &=2\pi+\frac{\pi}{4} \\ \cos \frac{18\pi}{8} &= \cos \left(2\pi+\frac{\pi}{4}\right) \\ &= \cos \frac{\pi}{4} \\ &= \frac{\sqrt{2}}{2} \end{align*}
We then have the half-angle formula $$ \cos \frac{1}{2}\theta = (-1)^{\lfloor (\theta + \pi) / (2\pi) \rfloor} \sqrt{\frac{1 + \cos \theta}{2}} $$ Setting $\theta = \frac{18\pi}{8}$, we have \begin{align*} \cos \left(\frac{1}{2}\right)\left(\frac{18\pi}{8}\right) &= (-1)^{\lfloor ((18\pi/8) + \pi) / (2\pi) \rfloor} \sqrt{\frac{1 + \frac{\sqrt{2}}{2}}{2}} \\ \cos \frac{9\pi}{8} &= (-1)^{\lfloor(13\pi/4)/(2\pi)\rfloor}\sqrt{\frac{2 + \sqrt{2}}{4}} \\ &= (-1)^{\lfloor 13\pi / 8\rfloor} \frac{\sqrt{2 + \sqrt{2}}}{2} \\ &= (-1)^1 \frac{\sqrt{2 + \sqrt{2}}}{2} \\ \cos \frac{9\pi}{8} &= - \frac{\sqrt{2 + \sqrt{2}}}{2} \end{align*}
$$ \cos(x)=\sin(2x)$$
$$ \Leftrightarrow 2\sin(x)\cos(x)-\cos(x)=0$$
$$ \Leftrightarrow \cos(x)(2\sin(x)-1)=0$$
$$ \Leftrightarrow \cos(x)=0 ~~~\sin(x)=\frac{1}{2}$$
$$ \Leftrightarrow x=\pm\frac{\pi}{2}+2\pi k ~~~ x=\pi-\frac{\pi}{6}+2k\pi $$
where $k \in Z$
Picture of intersections: 
Best Answer
When an angle has no unit, it is in radians. Since angle units are convertible, we may as well convert $\theta$ to radians – $\frac\pi2<\theta<\pi$ – and find that over that interval $\sin\frac2\theta$ is positive.