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ana321

  • 2 years ago

i need help

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  1. Trustted
    • 2 years ago
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    help in wat

  2. ana321
    • 2 years ago
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    use half angle formulas to rewrite \[\sqrt{1-\cos30 degrees/2}\]

  3. dumbcow
    • 2 years ago
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    \[\sin (\theta/2) = \sqrt{(1-\cos \theta)/2}\]

  4. ana321
    • 2 years ago
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    using that formula o but..

  5. dumbcow
    • 2 years ago
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    so you can rewrite it as sin(15)

  6. ana321
    • 2 years ago
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    hwo do i get there i cantsee

  7. dumbcow
    • 2 years ago
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    ??

  8. ana321
    • 2 years ago
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    what i mean the answer of the problem i sin 15 how do i get there id ont understand

  9. dumbcow
    • 2 years ago
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    theta = 30 so theta/2 = 30/2 = 15

  10. ana321
    • 2 years ago
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    can u show that

  11. dumbcow
    • 2 years ago
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    show what...prove that 30/2 = 15

  12. ana321
    • 2 years ago
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    no the whole process

  13. dumbcow
    • 2 years ago
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    well i looked up the half-angle identity for sin (theta/2) **shown above** it matches your expression except you have 30 degrees for the angle

  14. ana321
    • 2 years ago
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    ok ,..?

  15. ana321
    • 2 years ago
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    and

  16. ana321
    • 2 years ago
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    ok

  17. ana321
    • 2 years ago
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    nee help trig= rewrite this using half angle formula sqrt{1= \cos 135 degress/2}\]

  18. dumbcow
    • 2 years ago
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    is that a plus or minus ?

  19. ana321
    • 2 years ago
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    sorry let me rewrite it

  20. ana321
    • 2 years ago
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    \[rewrtite using half aangle formulas \sqrt{1+ \cos 135 dgrees/2}\]

  21. dumbcow
    • 2 years ago
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    plus means its the cosine half angle formula take the given angle --> 135 divide it by 2 --> 135/2 = 67.5 answer = cos(67.5)

  22. ana321
    • 2 years ago
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    so that my asnwer my teacher doesn tlike decimals

  23. dumbcow
    • 2 years ago
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    then write it as a fraction...the main thing is do you get the concept? find appropriate formula , then take half the angle

  24. ana321
    • 2 years ago
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    hello?

  25. ana321
    • 2 years ago
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    can u show me thewhole process and ow do is hwo it in fraction

  26. ana321
    • 2 years ago
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    hello

  27. ana321
    • 2 years ago
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    what wre u gonnaa say

  28. dumbcow
    • 2 years ago
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    are you serious...i showed you step by step...there is no other process you should have a reference in your book or notes showing the half-angle formulas the fraction would simply be 135/2 http://www.freemathhelp.com/trig-double-angles.html

  29. ana321
    • 2 years ago
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    need help trig rewrite using half angle idnetities =1-cos210 degress/sin210 degrees

  30. ana321
    • 2 years ago
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    need help trig rewrite using half angle idnetities =1-cos210 degress/sin210 degrees

  31. ana321
    • 2 years ago
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    heloo

  32. dumbcow
    • 2 years ago
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    \[\rightarrow \frac{1-\cos 210}{\sin 210}\]

  33. ana321
    • 2 years ago
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    yes i know the formula i dont know how to use it

  34. dumbcow
    • 2 years ago
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    please don;t message me the same thing you are posting, thanks and this is different it isn't any of the half-angle formulas

  35. ana321
    • 2 years ago
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    no yes

  36. dumbcow
    • 2 years ago
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    \[\frac{1-\cos 210}{\sin 210} = \csc 210 - \cot 210\] not sure how else to rewrite it unless you want to use sum/difference formulas

  37. ana321
    • 2 years ago
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    ok tanks

  38. ana321
    • 2 years ago
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    s o how do i ewrite using half angle idnetities =1-cos210 degress/sin210 degrees

  39. ana321
    • 2 years ago
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    by just saying it equals to 1-cos theta/sin theta

  40. ana321
    • 2 years ago
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    ???

  41. dumbcow
    • 2 years ago
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    umm i already answered you before regarding this question

  42. ana321
    • 2 years ago
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    no i dont grasp it

  43. dumbcow
    • 2 years ago
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    i don't think you can rewrite using half-angle formulas look for yourself...see that the expression does not match any the formulas

  44. ana321
    • 2 years ago
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    its what it says n my book

  45. ana321
    • 2 years ago
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    yes it does tan theta /2

  46. ana321
    • 2 years ago
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    hello

  47. dumbcow
    • 2 years ago
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    yes i am wrong...my ref sheet has a different but equivalent formula for tangent sorry...hey i learned something anyway, now just take half the angle

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