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anonymous

  • 5 years ago

use a double- angle identity to find the exact value sin22.5^ocos22.5^o

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  1. anonymous
    • 5 years ago
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    why I am still at 19 fans, after spending the last 5 hours working my *** out?

  2. anonymous
    • 5 years ago
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    if you help me with this i will become a fan

  3. anonymous
    • 5 years ago
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    ok, one min

  4. anonymous
    • 5 years ago
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    the double angle identities are as follows: \[\cos (A - B) = (\cos A)(\cos B) + (\sin A)(\sin B) \] \[\cos (A + B) = (\cos A)\cos B - (\sin A)\sin B \] \[\sin (A + B) = (\sin A)(\cos B) + (\cos A)(\sin B) \] \[\sin (A - B) = (\sin A)(\cos B) - (\cos A)(\sin B) \] alright andy :) you can lead from here ^_^

  5. anonymous
    • 5 years ago
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    hmm, am I mixing it all up with the double angle identities? those are the sin and cos indentities , hmm, please correct me if I'm wrong ^_^

  6. anonymous
    • 5 years ago
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    double angle identities go like this, egzample: sin(2x)=2sinxcosx

  7. nowhereman
    • 5 years ago
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    That is just the above with setting A=B

  8. anonymous
    • 5 years ago
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    no the identities are sin2A=2sineAcosA cos2A-cos^2A-sin^2A =1-2sin^2A =2cosA-1

  9. anonymous
    • 5 years ago
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    or right lol , thank you, alright angoo, give it a try ^_^

  10. anonymous
    • 5 years ago
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    ok I am bad at this, I guess I must have skipped school the day this stuff was tought

  11. anonymous
    • 5 years ago
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    give it a try :)

  12. anonymous
    • 5 years ago
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    yeah same here

  13. anonymous
    • 5 years ago
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    nah, stary, you go ahead, I dont get this stuff yet

  14. anonymous
    • 5 years ago
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    ask your beloved google lol

  15. anonymous
    • 5 years ago
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    well, Google can give me examples, and explanations, and I can learn from them, but it would take too much time... so just go ahead and do it mimi

  16. anonymous
    • 5 years ago
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    this is my new signature btw: ▓▒░╔ᴧᴨᴅᴙiᴜs╖░▒▓

  17. anonymous
    • 5 years ago
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    LOL

  18. anonymous
    • 5 years ago
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    alright LoveL , let me get this clear : you want to find the answer of : sin(22.5)cos(22.5)? or do you want to find the exact angle for sin(22.5)=? or cos(22.5)=? ^_^ lol, it's filled with mystery, but I like it :)

  19. nowhereman
    • 5 years ago
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    I will show you, first of all: \[\sin (22.5^\circ)\cos(22.5^\circ) = \sin(\frac{180^\circ}{8})\cos(\frac{180^\circ}{8}) = \sin(\frac{π}{8})\cos(\frac π 8)\]

  20. anonymous
    • 5 years ago
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    nvm, follow up with nowhereman ^_^ he'll lead you through.

  21. anonymous
    • 5 years ago
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    so, loveinglife, do you feel happy today? because I do, but I could become even more happy if you gift me a fan =D In that case, I will give you a fan too, so you become a lifesaver, and I become a superstar, a clear winwin =D

  22. anonymous
    • 5 years ago
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    LOL, and you're still on it

  23. anonymous
    • 5 years ago
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    @BecomeMyFan im your fan now. thank you for trying

  24. nowhereman
    • 5 years ago
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    Then \[\sin(\frac π 8)\cos(\frac π 8) = \frac 1 2 \cdot \left(2\sin(\frac π 8)\cos(\frac π 8)\right)\] Thus by applying that double arc formula on the inner part (in reverse direction if you want) you get \[=\frac 1 2 \sin(2\frac π 8) = \frac 1 2\sin(\frac π 4) = \frac 1 4 \sqrt 2\]

  25. anonymous
    • 5 years ago
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    Oh my god your awesome. thank you soo much you just saved my trig grade

  26. nowhereman
    • 5 years ago
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    You're welcome :-D

  27. anonymous
    • 5 years ago
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    love, are you in highschool?

  28. anonymous
    • 5 years ago
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    or in college

  29. anonymous
    • 5 years ago
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    ?

  30. anonymous
    • 5 years ago
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    because if you are doing this stuff in high school, I must have skipped ALOT of it

  31. anonymous
    • 5 years ago
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    im saddly in college and i thought i was good in math till i got here

  32. anonymous
    • 5 years ago
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    :)

  33. anonymous
    • 5 years ago
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    why am I still a mear star?

  34. anonymous
    • 5 years ago
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    I wanna be a superstar

  35. anonymous
    • 5 years ago
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    it was supposed to rhyme

  36. anonymous
    • 5 years ago
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    =D

  37. anonymous
    • 5 years ago
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    sorry cant help you there but im done, but i will probably be back at some point

  38. anonymous
    • 5 years ago
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    it is part of a poem I am writting about OpenStudy

  39. anonymous
    • 5 years ago
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    =)

  40. anonymous
    • 5 years ago
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    you write poems?

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