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Yahoo! Group Title

Prove that cos2 2x – cos2 6x = sin 4x sin 8x

  • one year ago
  • one year ago

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  1. waterineyes Group Title
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    Let us start from the start..

    • one year ago
  2. Yahoo! Group Title
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    it is cos^2...ok

    • one year ago
  3. Mimi_x3 Group Title
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    \[cos^{2}2x -cos^{2}6x = sin4xsin8x\]?

    • one year ago
  4. Yahoo! Group Title
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    yes

    • one year ago
  5. Mimi_x3 Group Title
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    have you tried cos^{2}2x = (1+cos4x)/2?

    • one year ago
  6. Yahoo! Group Title
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    no

    • one year ago
  7. apoorvk Group Title
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    \[\cos^2 A - \cos^2 B = \sin(A+B)\sin(B-A)\] This is a standard formula. So, we need a proof right?

    • one year ago
  8. Mitul Group Title
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    (1+cos4x)/2-(1+cos12x)/2 = (1+cos 4x+cos12x+cos4xcos12x)/2 and then defactorize the last term...and factorize the second and third term

    • one year ago
  9. Yahoo! Group Title
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    @apoorvk yup proof needed..

    • one year ago
  10. Mitul Group Title
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    @apoorvk it is cosA-cosBand not their squares for standard formula

    • one year ago
  11. waterineyes Group Title
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    \[\cos^2(2x) - \cos^2(6x) \implies (\cos(2x) - \cos(6x))(\cos(2x) + \cos(6x))\] So: Using: \[\cos(C) + \cos(D) = 2\cos(\frac{C+ D}{2})\cos(\frac{C-D)}{2})\] \[\cos(C) - \cos(D) = 2\sin(\frac{C+ D}{2})\sin(\frac{D-C}{2})\] \[(2\sin(4x) \sin(2x))(2\cos(4x)\cos(2x)) \implies (2\sin(2x)\cos(2x)) \cdot (2\sin(4x) \cos(4x))\] \[\implies \sin(4x) \cdot \sin(8x)\]

    • one year ago
  12. waterineyes Group Title
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    Using: \[2\sin(x)\cos(x) \implies \sin(2x)\]

    • one year ago
  13. waterineyes Group Title
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    I think above you by mistake written (B-A) @apoorvk

    • one year ago
  14. waterineyes Group Title
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    You are doing wrong I guess..

    • one year ago
  15. waterineyes Group Title
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    Rewrite for cos(A) - cos(B) = ??

    • one year ago
  16. Yahoo! Group Title
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    @apoorvk u told me it is sin(a+b).sin(b-a) but u have written differ

    • one year ago
  17. apoorvk Group Title
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    yeah I have memory problems -.- Amnesia *sigh* *sniff*

    • one year ago
  18. waterineyes Group Title
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    \[\cos(A) - \cos(B) = 2\sin(\frac{A+B}{2}) \cdot \sin(\frac{B-A}{2})\]

    • one year ago
  19. waterineyes Group Title
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    And not sin(a-B) in the last..

    • one year ago
  20. Yahoo! Group Title
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    @apoorvk if i use ur method we wont get that lol

    • one year ago
  21. apoorvk Group Title
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    NOOOOOOOO!!!!!!!!!!!!!!!!!! What have i been doing? /_\

    • one year ago
  22. apoorvk Group Title
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    Okay please tweak it and use it as needed, we know the procedure atleast. Or should I repost?

    • one year ago
  23. waterineyes Group Title
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    See: \[\cos^2(2x) - \cos^2(6x) = \sin(A+B)\sin(A-B) \implies \sin(2x - 6x)\sin(2x + 6x)\] Don't you think we get negative ??? @apoorvk

    • one year ago
  24. apoorvk Group Title
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    yeah i got it ate already - unfortunately i can't view my raw LateX codes (some stupid bug :\), so I'll have to edit it manually. NOOOOOO!!!!!!!!!

    • one year ago
  25. apoorvk Group Title
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    Sir what you think is correct, if you want a confirmation. -.-

    • one year ago
  26. waterineyes Group Title
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    There must be condition for doing this formula : I am discussing with you @apoorvk

    • one year ago
  27. waterineyes Group Title
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    The above said formula will work if B > A..

    • one year ago
  28. apoorvk Group Title
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    \[\cos^2A−\cos^2B\] \[=(\cos A+\cos B)(\cos A−\cos B)\] \[=[2\cos(\frac{A+B}2)\cos(\frac{A−B}2)][2\sin(\frac{A+B}2)\sin(\frac{B-A}2)]\] \[=[2\sin(\frac{A+B}2)\cos(\frac{A+B}2)][2\sin(\frac{B-A}2\cos(\frac{B-A}2)]\] \[=\Large\boxed{\sin(A+B)\sin(B-A)}\] Happy nitpickers? --____--

    • one year ago
  29. waterineyes Group Title
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    Do you think It will work ??

    • one year ago
  30. apoorvk Group Title
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    Water in eyes, Fish in Water \(\large \rightarrow\) so Fish in eyes. -_-

    • one year ago
  31. waterineyes Group Title
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    I am asking you @apoorvk

    • one year ago
  32. waterineyes Group Title
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    If fish is eyes then why are looking upwards?? Is fish in your eyes??

    • one year ago
  33. apoorvk Group Title
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    I paarsonaaally bilib it's Phaathar bill wa-ark! -,- (dish phormula)

    • one year ago
  34. waterineyes Group Title
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    What is this?? @apoorvk

    • one year ago
  35. apoorvk Group Title
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    U know aandarishtand? oowhaat u know no inglis?? BAH HUMBUG!

    • one year ago
  36. apoorvk Group Title
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    Inglish pipul go, liebe children behind -_-

    • one year ago
  37. apoorvk Group Title
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    'Addition', man, addition ----> " sin (A '+' B) - Y U LOVE SUBTRACTION??

    • one year ago
  38. waterineyes Group Title
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    Okay my mistake..

    • one year ago
  39. apoorvk Group Title
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    U inglish phail, I inglish oowhale. -____-

    • one year ago
  40. waterineyes Group Title
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    You failed in english? ok I got this.. Better luck next time.. @apoorvk

    • one year ago
  41. apoorvk Group Title
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    Hu shaid I phail in inglish?? U no si mai bandarfool inglish haan??? o.O I gets the 105 out of 100 -__- u get how many?

    • one year ago
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