sammietaygreen
  • sammietaygreen
Check answer?
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!
sammietaygreen
  • sammietaygreen
I think it's 0 but I feel like I'm wrong
sammietaygreen
  • sammietaygreen
@jim_thompson5910
tkhunny
  • tkhunny
Did we draw right triangles?

Looking for something else?

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

More answers

sammietaygreen
  • sammietaygreen
Yes, and tried graphing
jim_thompson5910
  • jim_thompson5910
using a calculator, what is the value of \[\Large \tan^{-1}\left(\frac{\sqrt{3}}{3}\right)\]
sammietaygreen
  • sammietaygreen
26.1?
jim_thompson5910
  • jim_thompson5910
make sure your calculator is in degree mode
jim_thompson5910
  • jim_thompson5910
btw you can type `arctan(sqrt(3)/3) in degrees` into google and google acts like a calculator. Tell me what result google says
jim_thompson5910
  • jim_thompson5910
that is if you aren't sure how to change from radians to degrees on your original calculator
sammietaygreen
  • sammietaygreen
30 degrees it says
jim_thompson5910
  • jim_thompson5910
yep, now compute the arcsine of 1/2
sammietaygreen
  • sammietaygreen
0.78 radians
jim_thompson5910
  • jim_thompson5910
remember: you need to be in degree mode
tkhunny
  • tkhunny
|dw:1450141910185:dw|
jim_thompson5910
  • jim_thompson5910
if you are using google, type in `arcsin(1/2) in degrees`. You have to put "in degrees" at the end to tell google to report the result in degree mode
sammietaygreen
  • sammietaygreen
still coming up 30 degrees
jim_thompson5910
  • jim_thompson5910
yes, so \[\Large {\color{blue}{\tan^{-1}\left(\frac{\sqrt{3}}{3}\right)}} = {\color{blue}{30^{\circ}}}\] \[\Large {\color{green}{\sin^{-1}\left(\frac{1}{2}\right)}} = {\color{green}{30^{\circ}}}\] which means \[\Large \cos\left({\color{blue}{\tan^{-1}\left(\frac{\sqrt{3}}{3}\right)}} + {\color{green}{\sin^{-1}\left(\frac{1}{2}\right)}}\right)\] \[\Large =\cos\left({\color{blue}{30^{\circ}}} + {\color{green}{30^{\circ}}}\right)\] \[\Large =\cos\left(60^{\circ}\right) = ???\]
sammietaygreen
  • sammietaygreen
1/2
jim_thompson5910
  • jim_thompson5910
correct
jim_thompson5910
  • jim_thompson5910
that's your final answer
sammietaygreen
  • sammietaygreen
Omg thank you. I don't even know how I got what I got

Looking for something else?

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