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treefriendly

  • 2 years ago

Can someone help me prove sin^2(x)cos^2(x) = (1-cos4x)/theta? I'm having trouble with trigonometry and I can't really grasp the material. Please and thank you

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  1. hartnn
    • 2 years ago
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    \(\cos 2x = 2\cos^2x-1 = 1-2\sin^2x\) so can you find cos^2x =..... ? and sin^2x =.... from this formula ?

  2. treefriendly
    • 2 years ago
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    Not really :/

  3. hartnn
    • 2 years ago
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    \(\cos 2x = 2\cos^2x-1 \) Add 1 to both sides, what u get ?

  4. treefriendly
    • 2 years ago
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    1 + cos2x = 2 cos^2 x?

  5. hartnn
    • 2 years ago
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    thats correct, now divide both sides by 2 what u get ?

  6. treefriendly
    • 2 years ago
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    (1+cos2x)/2 = cos^2 x?

  7. hartnn
    • 2 years ago
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    yes, correct. in the same way, isolate sin^2 x from \(\cos 2x = 1-2\sin^2x\)

  8. treefriendly
    • 2 years ago
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    so then cos2x-1 = -2sin^2 x?

  9. hartnn
    • 2 years ago
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    yes, now divide both sides by -2 what u get ?

  10. treefriendly
    • 2 years ago
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    (cos2x-1)/-2 = sin^2 x?

  11. treefriendly
    • 2 years ago
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    but I thought you couldn't divide by -2 because it's with the sine?

  12. hartnn
    • 2 years ago
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    yes, or you can say, (1-cos 2x)/2 = sin^2 x you can divide, there's no problem with that. now substitute both these results in left side of your equation sin^2x . cos^2x =... ?

  13. treefriendly
    • 2 years ago
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    what happened to the negative sign that was on the 2?

  14. hartnn
    • 2 years ago
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    cos2x -1 got turned into 1-cos2x because of that negative sign.

  15. treefriendly
    • 2 years ago
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    ohh okay. so ((1-cos^2 x)/2) ((1+cos2x)/2)) ?

  16. hartnn
    • 2 years ago
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    how cos^2 x ? aren't both of them cos 2x ?

  17. treefriendly
    • 2 years ago
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    Darn. Yeah, you're right. I was looking at something else when I was typing

  18. hartnn
    • 2 years ago
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    so, ((1-cos2 x)/2) ((1+cos2x)/2)) now use \((a-b)(a+b)=a^2+b^2\) so, (1-cos2x)(1+cos2x) =... ?

  19. treefriendly
    • 2 years ago
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    1+cos2x-cos2x-2cos^2 x

  20. hartnn
    • 2 years ago
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    check your last term. its not -2cos^2x

  21. treefriendly
    • 2 years ago
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    errr..... -cos^2 x^2? or -cos^2 2x?

  22. hartnn
    • 2 years ago
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    -cos^2 2x

  23. hartnn
    • 2 years ago
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    so, you have 1-cos^2 2x according to \(\cos 2x = 2\cos^2x-1\) what will be cos^2 2x=... ?)

  24. treefriendly
    • 2 years ago
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    cos2x?

  25. hartnn
    • 2 years ago
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    \(\cos 2x = 2\cos^2x-1 \\cos 2(2x) = 2\cos^2(2x)-1 \\ \cos^22x =(\cos 4x+1)/2 \) got this ?

  26. treefriendly
    • 2 years ago
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    kind of yeah

  27. hartnn
    • 2 years ago
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    so, whats \(\\~ \\ 1-\cos^22x =1-(\cos 4x+1)/2=..?\)

  28. treefriendly
    • 2 years ago
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    cos 2 (2x) = cos ^2 (2x) ?

  29. hartnn
    • 2 years ago
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    ou had \( ((1-\cos2 x)/2) (1+\cos2x)/2))= (1-\cos^2 2x)/4 \\ =(2-\cos 4x-1)=1-\cos 4x\)

  30. AravindG
    • 2 years ago
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    @treefriendly got it now?

  31. treefriendly
    • 2 years ago
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    Sort of, but I'm a bit lost.

  32. AravindG
    • 2 years ago
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    can you tell me where?

  33. treefriendly
    • 2 years ago
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    The last part. What happened to the 8 from the initial equation?

  34. AravindG
    • 2 years ago
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    i dont see any 8 here

  35. treefriendly
    • 2 years ago
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    Sorry, it's getting late and I've been misreading

  36. AravindG
    • 2 years ago
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    i see ..then i suggest you take rest do it afresh the next day

  37. treefriendly
    • 2 years ago
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    I meant what happened to theta that was being divided from 1-cos4x?

  38. AravindG
    • 2 years ago
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    there is no theta here

  39. treefriendly
    • 2 years ago
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    prove sin^2(x)cos^2(x) = (1-cos4x)/theta?

  40. hartnn
    • 2 years ago
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    its not theta, its 8 \(((1-\cos2 x)/2) (1+\cos2x)/2))= (1-\cos^2 2x)/4 \\\\=(1-(\cos4x+1)/2)/4\\ =(2-\cos 4x-1)/8=(1-\cos 4x)/8\)

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