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AravindG Group Title

how to find value of sin 18

  • 2 years ago
  • 2 years ago

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  1. Argonx16 Group Title
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    Take a calculator, and hit the "sin" button and type 18 in. You (may) need a scientific/graphing/google calculator. If you are asking how to do so by hand, I am not qualified to tell you.

    • 2 years ago
  2. AravindG Group Title
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    there must be a method using the formulas

    • 2 years ago
  3. AravindG Group Title
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    turing??

    • 2 years ago
  4. TuringTest Group Title
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    I am thinking....

    • 2 years ago
  5. SUROJ Group Title
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    (18*5)=90

    • 2 years ago
  6. AravindG Group Title
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    so?

    • 2 years ago
  7. TuringTest Group Title
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    all I got is the fact that we can write this as\[\sin(\frac\pi{10})\]thinking if we can go somewhere from there...

    • 2 years ago
  8. TuringTest Group Title
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    looking at some way to write this such that we can apply an identity to it...

    • 2 years ago
  9. SUROJ Group Title
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    http://answers.yahoo.com/question/index?qid=20061025091609AASJNUn

    • 2 years ago
  10. AravindG Group Title
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    whoa!!

    • 2 years ago
  11. TuringTest Group Title
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    why is \[\sin(3x)=\cos(2x)\]?

    • 2 years ago
  12. SUROJ Group Title
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    Most of peoples are not smart....they are clever

    • 2 years ago
  13. campbell_st Group Title
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    use the power series expansion \[\sin(x) = x/1! - x^3/3! + x^5/5! -x^7/7!.....\]

    • 2 years ago
  14. AravindG Group Title
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    hw put valu 18>??

    • 2 years ago
  15. AravindG Group Title
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    bt it goes on

    • 2 years ago
  16. SUROJ Group Title
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    if x= 18, 2x+3x=90

    • 2 years ago
  17. AravindG Group Title
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    Another way to do this is to start with a regular pentagon where the length of each side is one unit. I'll outline the steps. (You can fill in the details.) Call the pentagon ABCDE. Draw diagonals AC and BE, which intersect at point F. Using the fact that each angle of the pentagon is 108 degrees, you can prove the triangles BFC and AFE are isosceles. Therefore the lengths of line segments CF and EF are also 1 unit. Also note the angles for the triangles are 72-72-36. Now show that ABF and CAB are similar triangles. Since they are similar, this means BE/AB = AB/BF. Since BE = EF + BF, substitute to get (EF + BF)/AB = AB/BF. Let's define x to be the length of BF. Then, since EF = 1, the equation becomes (1 + x)/1 = 1/x, or x^2 + x - 1 = 0. Use the quadratic equation to get x = (sqrt(5) - 1) / 2. Finally, let's create an 18-72-90 by drawing the altitude of triangle BFC from vertex C to side BF. Since BC = FC, the altitude hits BF at the midpoint of BF, which we will call point G. So, (a) triangle GBE is an 18-72-90 triangle, (b) BG = BF/2 = x/2. Using this 18-72-90 triangle, sin 18 degrees = BG / BC = x/2. Therefore sin 18 = (sqrt(5) - 1) / 4.

    • 2 years ago
  18. AravindG Group Title
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    i liked this

    • 2 years ago
  19. SUROJ Group Title
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    2x=90-3x

    • 2 years ago
  20. SUROJ Group Title
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    cos both side

    • 2 years ago
  21. AravindG Group Title
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    k

    • 2 years ago
  22. SUROJ Group Title
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    cos(2x)=sin(3x)

    • 2 years ago
  23. AravindG Group Title
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    thx evryone

    • 2 years ago
  24. SUROJ Group Title
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    catch it?

    • 2 years ago
  25. TuringTest Group Title
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    gotchya, thx

    • 2 years ago
  26. AravindG Group Title
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    campbell pls tell ur method

    • 2 years ago
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