anonymous
  • anonymous
What is the exact value of cos 17pi/8? a) square root of 2+the square root of 2/4 b.) 0.38 c.) 0.99 d.)square root of 2-the square root of 2/4 ***My Answer: D***
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.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
Loser66
  • Loser66
Nope
anonymous
  • anonymous
Dangit hahaha. My second choice would have to be A then
Loser66
  • Loser66
Nope again, ahhahha

Looking for something else?

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

More answers

anonymous
  • anonymous
Dangit(lol)...Not the best at trig identities. Sorry
Loser66
  • Loser66
ok, hint 17pi/8 = 16pi/8 + pi/8
Loser66
  • Loser66
and you have the perfect angle 16pi/8 = 2pi
Loser66
  • Loser66
break it out by cos (a+b) =....
anonymous
  • anonymous
How would I find the a and b?
anonymous
  • anonymous
Would it just be a= 16pi/8 and b= pi/8?
Loser66
  • Loser66
hey, \(cos (\dfrac{17\pi}{8}=cos (\dfrac{16\pi +\pi}{8}\)
Loser66
  • Loser66
yup
anonymous
  • anonymous
Oh I see. Ok
anonymous
  • anonymous
So from there we get it into the form...\[\cos(\frac{ 16\pi }{ 8 })\cos(\frac{ \pi }{ 8 }) + \sin(\frac{ 16\pi }{ 8 })\sin(\frac{ \pi }{ 8 })\]
anonymous
  • anonymous
I think
anonymous
  • anonymous
@Loser66 Now this is the part that I have trouble understanding. I do not know where to go from here
Loser66
  • Loser66
whyyyyyyyyy? 16pi/8 = 2pi, right?
anonymous
  • anonymous
Yes
Loser66
  • Loser66
cos 2pi =1, right?
Loser66
  • Loser66
sin 2pi =0, ok?
anonymous
  • anonymous
Ok
Loser66
  • Loser66
so, at the end, you just have cos (pi/8 ) =0.99999
anonymous
  • anonymous
Ohhhhhh ok! That was my problem all along! I would divide pi/8 without the cos! Haha Thank you so much!
Loser66
  • Loser66
ok
anonymous
  • anonymous
What the... my computer just told me that it was A :(
Loser66
  • Loser66
Facepalm. hehehe... let's get other's help @campbell_st
mathstudent55
  • mathstudent55
\(\dfrac{17\pi}{8} = \dfrac{16\pi}{8} + \dfrac{\pi}{8} = 2\pi + \dfrac{\pi}{8}\) \(\cos \dfrac{17\pi}{8} = \cos 2\pi + \dfrac{\pi}{8} = \cos \dfrac{\pi}{8} \) \(\Large \cos \dfrac{\pi}{8} = \cos \left ( {\dfrac{\frac{\pi}{4}}{2} } \right)\) \(\cos \dfrac{\theta}{2} = \sqrt{\dfrac{1}{2}(1 + \cos \theta) }\) \(\Large \cos \dfrac{\pi}{8} = \cos \left ( {\dfrac{\frac{\pi}{4}}{2} } \right) = \sqrt{\dfrac{1}{2}(1 + \cos \dfrac{\pi}{4})}\) \(= \sqrt{\dfrac{1}{2} (1 + \dfrac{\sqrt{2}}{2})} = {\sqrt{\dfrac{1}{2} + \dfrac{\sqrt{2}}{4}}} \) \(= \sqrt{\dfrac{2 + \sqrt 2}{4}} = \dfrac{\sqrt{2 + \sqrt 2 } }{ 2 }\)
Loser66
  • Loser66
WWWWWWWWWWWWoooooah!!! It is much.... more logic than my way. Thank you so much @mathstudent55
mathstudent55
  • mathstudent55
You're welcome.
mathstudent55
  • mathstudent55
@Loser66 You were trying to use the identity for the cosine of a sum of angles. That method does not work in this problem. That method works if you can break up the angle into two angles whose cosines you know. The problem here is that the cos of pi/8 is not one of the know values. On the other hand, the cos of pi/4 is well known, so using the identity of the cosine of a half angle gives you the result.

Looking for something else?

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