kalli
  • kalli
Use the half-angle formula to evaluate tan(17pi/12)
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
freckles
  • freckles
well \[\tan(\frac{17\pi}{12}) \\ y=\tan(x) \text{ has period } \pi \\ \tan(\frac{17\pi}{12}-\pi)=\tan(\frac{17\pi}{12}-\frac{12\pi}{12})=\tan(\frac{5\pi}{12}) \\ \text{ now } 2 \cdot \frac{5 \pi}{12}=\frac{5\pi}{6} \\ \text{ so we have that we want to use the half-angle formula on } \\ \tan( \frac{1}{2} \cdot \frac{5 \pi}{6})\]
freckles
  • freckles
do you know the half angle identity for tan?
kalli
  • kalli
no, only for sin

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

freckles
  • freckles
so you can't use the half angle identity for tan?
freckles
  • freckles
you can only use the one for sin?
freckles
  • freckles
the weird thing is you have tan here not sin
freckles
  • freckles
or do you mean you can use it but you just don't know it?
freckles
  • freckles
http://www.purplemath.com/modules/idents.htm do you see half-angle identities on this page?
freckles
  • freckles
notice the three different ones they wrote for tan(x/2)
freckles
  • freckles
use one of them
kalli
  • kalli
im confused because I don't really know how to do this lol
freckles
  • freckles
well first let's just copy of of the half-angle identities over from that page \[\tan(\frac{1}{2}x)=\frac{\sin(x)}{1+\cos(x)} \\ \text{ and we want to evaluate } \\ \tan(\frac{1}{2} \frac{5\pi}{6})=?\] do you see what to replace x with ?
freckles
  • freckles
compare tan(1/2*5pi/6) to tan(1/2*x)
freckles
  • freckles
x has to be 5pi/6 right?
freckles
  • freckles
so replace the x's in the identity with 5pi/6
kalli
  • kalli
ok one sec
kalli
  • kalli
tan(1/2*5pi/6)= sin(5pi/6) / 1+cos(5pi/6) ??
freckles
  • freckles
ok and ( ) around the bottom
freckles
  • freckles
so sin(5pi/6)=? and cos(5pi/6)=?
kalli
  • kalli
1/2 and -sqrt 3/2 ??
freckles
  • freckles
\[\tan(\frac{1}{2} \cdot \frac{5\pi}{6})=\frac{\frac{1}{2}}{1+\frac{-\sqrt{3}}{2}}\] multiply top and bottom by 2
freckles
  • freckles
\[\frac{\frac{2}{2}}{2+\frac{2(-\sqrt{3})}{2}}\]
freckles
  • freckles
can you simplify from there?
freckles
  • freckles
you know 2/2=1
kalli
  • kalli
Yes one second
kalli
  • kalli
2+ sqrt3
kalli
  • kalli
@freckles is that my answer?
freckles
  • freckles
let me check one sec
freckles
  • freckles
\[\frac{\frac{2}{2}}{2+\frac{2(-\sqrt{3})}{2}} \\ \frac{1}{2-\sqrt{3} } \\ \frac{1}{2 -\sqrt{3}} \cdot \frac{2 +\sqrt{3}}{2+\sqrt{3}} \\ \frac{2+\sqrt{3}}{4-3} \\ \frac{2 +\sqrt{3}}{1} \\ 2+\sqrt{3}\]
freckles
  • freckles
yep looks fabulous
kalli
  • kalli
thanks so much!!!
freckles
  • freckles
np

Looking for something else?

Not the answer you are looking for? Search for more explanations.