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AravindG

  • 4 years ago

wel i hav some doubts on trigonometruy wen i did revision can anyone help

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  1. AravindG
    • 4 years ago
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    what is the formula for cos 2A *cos 4A *cos 8A ETC

  2. anonymous
    • 4 years ago
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    cos2A=1-sinA i think..

  3. anonymous
    • 4 years ago
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    no

  4. anonymous
    • 4 years ago
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    meant 1-sin2A

  5. anonymous
    • 4 years ago
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    oh, no? ok

  6. AravindG
    • 4 years ago
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    11+cos2a is 2sin^2A

  7. ash2326
    • 4 years ago
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    cos 2A= cos^2 A- sin ^2 A

  8. anonymous
    • 4 years ago
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    \[\cos^2A=1-\sin^2A\] but it's not suqared

  9. AravindG
    • 4 years ago
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    .........

  10. anonymous
    • 4 years ago
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    no formula for this http://www.wolframalpha.com/input/?i=simplify+cos+2A+*cos+4A+*cos+8A

  11. ash2326
    • 4 years ago
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    aravind do want the simplification of cos 2A* cos 4A * cos 8A

  12. ash2326
    • 4 years ago
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    *you

  13. AravindG
    • 4 years ago
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    no i think there is a formula

  14. AravindG
    • 4 years ago
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    for n terms

  15. AravindG
    • 4 years ago
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    somewat like sin 2^n alpha/(2^n sin alpha)

  16. anonymous
    • 4 years ago
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    ah it doesn't end idk i haven't learn such formula in trig course

  17. AravindG
    • 4 years ago
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    its there tomas

  18. anonymous
    • 4 years ago
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    you are from different country...

  19. AravindG
    • 4 years ago
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    :P but math is universal :P

  20. anonymous
    • 4 years ago
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    but we don't learn same things

  21. anonymous
    • 4 years ago
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    so as i said i didn't learn it in my trig course

  22. AravindG
    • 4 years ago
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    k

  23. ash2326
    • 4 years ago
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    aravind I'm trying to think, you're right that there is a formula, but i'm not able to recall

  24. AravindG
    • 4 years ago
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    oh thn next doubt u find it and post here later

  25. AravindG
    • 4 years ago
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    u knw solutions of triangles?

  26. ash2326
    • 4 years ago
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    not very much but I'll try,

  27. AravindG
    • 4 years ago
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    :O

  28. AravindG
    • 4 years ago
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    in a triangle (a+b+c)(b+c-a)=kbc if 0<k<4 prove

  29. AravindG
    • 4 years ago
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    ??

  30. AravindG
    • 4 years ago
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    ash

  31. ash2326
    • 4 years ago
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    yeah aravind i'm thinking

  32. Diyadiya
    • 4 years ago
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    \[Cos2x=1-2\sin^2x\]

  33. Diyadiya
    • 4 years ago
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    \[Cos2x=\cos^2x-\sin^2x\]

  34. Diyadiya
    • 4 years ago
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    \[Cos2x=2\cos^2x-1\]

  35. ash2326
    • 4 years ago
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    yeah, i can help with you second question equation 1....(a+b+c)(b+c-a)= b^2+c^2-a^2+2bc now we know cosine formula a^2=b^2+c^2-2bc*cos A so here b^2+c^2-a^2=2bc cos A, subsitute this in equation 1 so 2bc cos A+ 2bc 2bc( cos A+1) now cos A 's max value is 1 and minimum 0 so the left side can be 0 or 4bc so k can vary from 0 to 4 0<k<4

  36. AravindG
    • 4 years ago
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    wow perfect

  37. vishal_kothari
    • 4 years ago
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    http://www.mediafire.com/?t2tdkmznnzq

  38. vishal_kothari
    • 4 years ago
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    handbook of mathematics...

  39. vishal_kothari
    • 4 years ago
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    it's not mine habit...

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