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anonymous

  • one year ago

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***

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  1. Loser66
    • one year ago
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    Nope

  2. anonymous
    • one year ago
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    Dangit hahaha. My second choice would have to be A then

  3. Loser66
    • one year ago
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    Nope again, ahhahha

  4. anonymous
    • one year ago
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    Dangit(lol)...Not the best at trig identities. Sorry

  5. Loser66
    • one year ago
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    ok, hint 17pi/8 = 16pi/8 + pi/8

  6. Loser66
    • one year ago
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    and you have the perfect angle 16pi/8 = 2pi

  7. Loser66
    • one year ago
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    break it out by cos (a+b) =....

  8. anonymous
    • one year ago
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    How would I find the a and b?

  9. anonymous
    • one year ago
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    Would it just be a= 16pi/8 and b= pi/8?

  10. Loser66
    • one year ago
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    hey, \(cos (\dfrac{17\pi}{8}=cos (\dfrac{16\pi +\pi}{8}\)

  11. Loser66
    • one year ago
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    yup

  12. anonymous
    • one year ago
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    Oh I see. Ok

  13. anonymous
    • one year ago
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    So from there we get it into the form...\[\cos(\frac{ 16\pi }{ 8 })\cos(\frac{ \pi }{ 8 }) + \sin(\frac{ 16\pi }{ 8 })\sin(\frac{ \pi }{ 8 })\]

  14. anonymous
    • one year ago
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    I think

  15. anonymous
    • one year ago
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    @Loser66 Now this is the part that I have trouble understanding. I do not know where to go from here

  16. Loser66
    • one year ago
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    whyyyyyyyyy? 16pi/8 = 2pi, right?

  17. anonymous
    • one year ago
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    Yes

  18. Loser66
    • one year ago
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    cos 2pi =1, right?

  19. Loser66
    • one year ago
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    sin 2pi =0, ok?

  20. anonymous
    • one year ago
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    Ok

  21. Loser66
    • one year ago
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    so, at the end, you just have cos (pi/8 ) =0.99999

  22. anonymous
    • one year ago
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    Ohhhhhh ok! That was my problem all along! I would divide pi/8 without the cos! Haha Thank you so much!

  23. Loser66
    • one year ago
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    ok

  24. anonymous
    • one year ago
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    What the... my computer just told me that it was A :(

  25. Loser66
    • one year ago
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    Facepalm. hehehe... let's get other's help @campbell_st

  26. mathstudent55
    • one year ago
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    \(\dfrac{17\pi}{8} = \dfrac{16\pi}{8} + \dfrac{\pi}{8} = 2\pi + \dfrac{\pi}{8}\) \(\cos \dfrac{17\pi}{8} = \cos 2\pi + \dfrac{\pi}{8} = \cos \dfrac{\pi}{8} \) \(\Large \cos \dfrac{\pi}{8} = \cos \left ( {\dfrac{\frac{\pi}{4}}{2} } \right)\) \(\cos \dfrac{\theta}{2} = \sqrt{\dfrac{1}{2}(1 + \cos \theta) }\) \(\Large \cos \dfrac{\pi}{8} = \cos \left ( {\dfrac{\frac{\pi}{4}}{2} } \right) = \sqrt{\dfrac{1}{2}(1 + \cos \dfrac{\pi}{4})}\) \(= \sqrt{\dfrac{1}{2} (1 + \dfrac{\sqrt{2}}{2})} = {\sqrt{\dfrac{1}{2} + \dfrac{\sqrt{2}}{4}}} \) \(= \sqrt{\dfrac{2 + \sqrt 2}{4}} = \dfrac{\sqrt{2 + \sqrt 2 } }{ 2 }\)

  27. Loser66
    • one year ago
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    WWWWWWWWWWWWoooooah!!! It is much.... more logic than my way. Thank you so much @mathstudent55

  28. mathstudent55
    • one year ago
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    You're welcome.

  29. mathstudent55
    • one year ago
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    @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.

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