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anonymous

  • 4 years ago

lim as x - -> 0 (1-cos5xcos7x)/((sin11x)^@)

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  1. anonymous
    • 4 years ago
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    answer is 0

  2. anonymous
    • 4 years ago
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    let me check it once more

  3. anonymous
    • 4 years ago
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    i want to know how did u find out the answer???please!

  4. dumbcow
    • 4 years ago
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    what is "^@"

  5. anonymous
    • 4 years ago
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    I am sorry, the answer is 37 / 121, but the algebra is tiresome

  6. anonymous
    • 4 years ago
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    give me some time to post the full solution.

  7. anonymous
    • 4 years ago
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    Since, Cos C + Cos D = 2.cos ((C + D) / 2)cos ((C -D) / 2) \[1 - \cos5x.\cos7x = 1 - (1/2) (\cos12x + \cos 2x)\] = 1/2(1 - COS12X) + 1/2 (1 - COS2X) = (1/2).2(SIN6X)^2 + (1/2).2(SINX)^2 = (SIN6X) ^ 2 + (SINX)^2 Hence the equation becomes -\[\sin ^{2}6x / \sin ^{2}11x + \sin ^{2}x/\sin ^{2}11x\] The first part \[= [(\sin ^{2}6x / (6x)^{2}) / (\sin ^{2}11x / (11x) ^{2})] (36 /121)\] Did you understand this part? (36 / 121 is introduced to nullify (6x)^2 and (11x)^2) Hence the first part becomes 36 / 121 Similarly the second part becomes 1 / 121 Hence it becomes in total = 37 / 121

  8. dumbcow
    • 4 years ago
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    yep, L'Hopitals Rule twice ---> limit = 37/121

  9. anonymous
    • 4 years ago
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    yeah, that is much easier :)

  10. dumbcow
    • 4 years ago
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    nice solution

  11. anonymous
    • 4 years ago
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    I will not forget L'Hospital anymore

  12. anonymous
    • 4 years ago
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    understood all part you had written. thank you very much for your help.

  13. anonymous
    • 4 years ago
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    do u have anything solution?

  14. anonymous
    • 4 years ago
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    I am sorry, but could not understand your question :), seemingly I dont have solution to this question

  15. anonymous
    • 4 years ago
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    ((sin11x)^2)

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