Answer:
Terminal points
[tex](x,y) = (\frac{1}{\sqrt{2} }) , (\frac{-1}{\sqrt{2} })[/tex]
Step-by-step explanation:
According to the question, it is provided that
[tex]\theta = \frac{7\pi}{4}[/tex]
Now
[tex]x = 1. cos (\frac{7\pi}{4})\\\\ = cos (2\pi - \frac{\pi}{4})\\\\= cos \frac{\pi}{4} \\\\= \frac{1}{\sqrt{2} }[/tex]
[tex]y = 1. sin (\frac{7\pi}{4})\\\\ = sin (2\pi - \frac{\pi}{4})\\\\= -sin \frac{\pi}{4} \\\\= \frac{-1}{\sqrt{2} }[/tex]
Now
[tex](x,y) = (\frac{1}{\sqrt{2} }) , (\frac{-1}{\sqrt{2} })[/tex]
These two represents the terminal points
We simply applied the above equations and then equate these two equations to determine the terminal points and therefore the same is to be considered
It could be figure out by using the x and y points
Therefore the two shows the terminal points
