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mitchelsewbaran

Verify the identity. cos 4u= cos^2(2u)- sin^2 (2u)

  • one year ago
  • one year ago

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  1. mathslover
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    Put 'u' as any angle ... let it be \(\large{\frac{\pi}{2}}\) .

    • one year ago
  2. mathslover
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    What do you get now as : \(\large{\cos(4(\frac{\pi}{2}))}\) = ?

    • one year ago
  3. mathslover
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    Can you tell me @mitchelsewbaran

    • one year ago
  4. .Sam.
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    I would use cos(4u) cos(2u+2u) Then by identity: cos(A+B)=cosAcosB-sinAsinB cos(2u)cos(2u)-sin(2u)sin(2u) cos^2(2u)-sin^2(2u)

    • one year ago
  5. mathslover
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    Is that verification or proof @.Sam. I think we have to verify and not to prove

    • one year ago
  6. mitchelsewbaran
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    cos(4(π 2 )) = 1

    • one year ago
  7. .Sam.
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    If they both equal then that's verifying too

    • one year ago
  8. mathslover
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    right now solve this : \[\large{\cos^2(2(\frac{\pi}{2}))- \sin^2(2(\frac{\pi}{2}))}\]

    • one year ago
  9. mitchelsewbaran
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    hold on

    • one year ago
  10. mathslover
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    take your time

    • one year ago
  11. mitchelsewbaran
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    1?

    • one year ago
  12. mathslover
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    yes can you tell me how you got that ?

    • one year ago
  13. mitchelsewbaran
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    I replaced that whole equation with cos(2pi). Taking the cosine of 2pi, I got 1

    • one year ago
  14. mathslover
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    o.O a small mistake @mitchelsewbaran : see here: \[\large{\cos^2(2(\frac{\pi}{2})) - \sin^2(2(\frac{\pi}{2}))}\] \[\large{\cos^2(\pi) - \sin^2(\pi)}\] \[\large{1-0= 1}\]

    • one year ago
  15. mitchelsewbaran
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    oh srry about that

    • one year ago
  16. mathslover
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    No problem I just corrected you and I hope that you will not repeat that again. Best of luck :)

    • one year ago
  17. mitchelsewbaran
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    I have only 2 more that I need to verify. I was wondering if you can help?

    • one year ago
  18. tkhunny
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    You cannot substitute a single value (or any finite number of values) to PROVE this. If you find one that doesn't work, that would be sufficient to DISprove it.

    • one year ago
  19. mathslover
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    Did you try yourself first?

    • one year ago
  20. mitchelsewbaran
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    can u help me verify these last 2?

    • one year ago
  21. mathslover
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    there is not fight @mitchelsewbaran . sorry if you felt bad. @tkhunny the question is to verify .. and hence we have to put a value and verify it

    • one year ago
  22. mitchelsewbaran
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    @mathslover ok :)

    • one year ago
  23. mitchelsewbaran
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    no I didn't feel bad @mathslover

    • one year ago
  24. tkhunny
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    @mathslover Demonstrating one or two values is just not what it is asking. It must be verified for ALL values. Since you cannot enter infinitely many values, there must be other methods employed. Demonstrating \(x = \pi\) says nothing of \(x = \sqrt{2}\)

    • one year ago
  25. mathslover
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    Ok @mitchelsewbaran post the one question of that two there in "ask a question forum" and surely I will help you but the second one will be solved by you and I will check it... if you get any problem in any concept you can ask me.

    • one year ago
  26. mitchelsewbaran
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    ok :)

    • one year ago
  27. mitchelsewbaran
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    @mathlover u're awesome, bro :)

    • one year ago
  28. mathslover
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    :)

    • one year ago
  29. mitchelsewbaran
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    ;)

    • one year ago
  30. mathslover
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    well I think @tkhunny has a point but still it is not given to verify for all values... it is just to verify.. but if we still go on for a proving method .. then it's surely not verifying

    • one year ago
  31. tkhunny
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    No, this is just not the case. "Verify the Identity" has meant to provide a general proof for all possible values in the entire Domain - at least since 1972. I am familiar with no reference in any mathematical text that has ever meant anything else. If only a few values were to be sufficient, the problem statement would say something like this, "Verify a few values and formulate a conjecture on whether or not this statement is generally true." "Verify the Identity" means ALL of them - leaving nothing out.

    • one year ago
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