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itsmichelle29

  • one year ago

help plz medal and fan Simplify sin^2 theta /1- cos^2 theta

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  1. Nnesha
    • one year ago
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    \[\huge\rm cos^2 \theta + \sin^2 \theta =1 \]you should remember this equation!!

  2. Nnesha
    • one year ago
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    solve for sin^2theta

  3. anonymous
    • one year ago
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    -cos(2 theta) 2 sin^2(theta)-1

  4. itsmichelle29
    • one year ago
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    cos^2 theta .. yes ????

  5. Nnesha
    • one year ago
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    what about it ? :=)

  6. itsmichelle29
    • one year ago
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    is that the answer ????

  7. Nnesha
    • one year ago
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    no

  8. itsmichelle29
    • one year ago
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    Okay let me try and get it

  9. Nnesha
    • one year ago
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    sure!

  10. imqwerty
    • one year ago
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    sin^2theta +cos^2theta = 1 1-cos^2theta=sin^2theta so we can write -> sin^2theta/sin^2theta = 1 but the condition is that sin(theta) does not equal 0

  11. itsmichelle29
    • one year ago
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    sin2theta

  12. itsmichelle29
    • one year ago
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    Yes ??

  13. Nnesha
    • one year ago
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    that's one of the option....

  14. imqwerty
    • one year ago
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    can u explain how did u get sin2theta

  15. itsmichelle29
    • one year ago
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    By 1-cos2theta=sin2theta........................... i think

  16. Nnesha
    • one year ago
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    \[\frac{ \sin^2\theta }{\color{ReD}{ 1-\cos^2 }}\] now replace 1-cos^2for sin^2x bec 1-cos^2x=sin^2x

  17. itsmichelle29
    • one year ago
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    Omg i dont understand

  18. Nnesha
    • one year ago
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    \[\huge\rm cos^2 \theta + \sin^2 \theta =1 \] solve for sin^2\[\rm sin^2\theta=1-\cos^2\theta\] you will get sin^2 =1-cos^2 right ?

  19. itsmichelle29
    • one year ago
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    Yes

  20. Nnesha
    • one year ago
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    okay so if sin^2theta =1-cos^2 you can replace the (1-cos^2) which is at the dneominator with sin^2 theta right ?

  21. Nnesha
    • one year ago
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    \[\huge\rm \frac{ \sin^2\theta }{\color{ReD}{ 1-\cos^2 }}\] replace 1-cos^2for sin^2x bec \[\huge\rm \color{reD}{ 1-\cos^2x}=\sin^2x \]

  22. itsmichelle29
    • one year ago
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    so the answer is sin right

  23. Nnesha
    • one year ago
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    \[\frac{ \sin^2 x }{ \color{reD}{\sin^2x}}=?\]

  24. Nnesha
    • one year ago
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    let sin^2x = a so a/a = ??

  25. Nnesha
    • one year ago
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    a divide by a = ??

  26. itsmichelle29
    • one year ago
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    0

  27. Nnesha
    • one year ago
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    no never!

  28. Nnesha
    • one year ago
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    there is always invisible one!!

  29. Nnesha
    • one year ago
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    2/2 = ??

  30. itsmichelle29
    • one year ago
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    1 lol

  31. Nnesha
    • one year ago
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    ye.-.

  32. itsmichelle29
    • one year ago
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    okay thnks

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