Answer:
[tex]\displaystyle -\sqrt{3} - 2[/tex]
Explanation:
If you recall the unit circle [from the polar graph], you would have no trouble at all figuring this out, but sinse you have trouble, do not worry about it. So, here is what you should have realised:
[tex]\displaystyle \boxed{1\frac{1}{2}\pi} = \frac{18}{12}\pi[/tex]
If you did not notise, [tex]\displaystyle 1\frac{7}{12}\pi,[/tex]or [tex]\displaystyle \frac{19}{12}\pi,[/tex]is just [tex]\displaystyle \frac{\pi}{12}[/tex]more than [tex]\displaystyle 1\frac{1}{2}\pi,[/tex]which means the exact value will either be a radical, followed by an integer or just a radical. In this case, accourding to the unit circle, you will have a radical, followed by an integer, and that will be this:
[tex]\displaystyle \frac{sin\:1\frac{7}{12}\pi}{cos\: 1\frac{7}{12}\pi} = tan\:1\frac{7}{12}\pi \\ \\ \boxed{-\sqrt{3} - 2} = \frac{sin\:1\frac{7}{12}\pi}{cos\: 1\frac{7}{12}\pi}[/tex]
It is all about memorisation of the unit circle, which I know is difficult, but you will get used to it soon.
*Now, if you had to find [tex]\displaystyle cos\:\frac{7}{12}\pi,[/tex]then the exact value would be the OPPOCITE of the exact value of [tex]\displaystyle sin\:\frac{7}{12}\pi,[/tex]which is [tex]\displaystyle -\frac{\sqrt{2} + \sqrt{6}}{4},[/tex]because you would be crossing into the second quadrant where the x-coordinates are negative, accourding to both the cartesian and polar graphs. In addition, sinse you ALREADY have both values for sine and cosine [accourding to your PREVIOUS QUESTION ASKED], all you need to do is divide both values in the order given above to arrive at the exact value for [tex]\displaystyle tan\:1\frac{7}{12}\pi.[/tex]
I am joyous to assist you at any time.
