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anonymous

  • one year ago

Use the squared Identities to simplify 2cos^2 (x) * cos^2 (x)

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  1. anonymous
    • one year ago
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    @ganeshie8 @JackJordan

  2. anonymous
    • one year ago
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    uhmm... how simple do you want it I mean an easy one would be just 2cos^4(x) lol

  3. anonymous
    • one year ago
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    a) 1-cos(4x) / 4 b) 1+4cos(2x)+cos(4x) / 4 c) 1+cos(4x) / 4 d) 1+4cos(2x)-cos(4x) / 4

  4. anonymous
    • one year ago
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    @ganeshie8

  5. anonymous
    • one year ago
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    are you sure you haven't miswritten the second one?

  6. anonymous
    • one year ago
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    b) 1+4cos(2x)+cos(4x) / 4

  7. anonymous
    • one year ago
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    thats what i have

  8. anonymous
    • one year ago
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    @ganeshie8

  9. anonymous
    • one year ago
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    @JackJordan

  10. anonymous
    • one year ago
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    @jim_thompson5910

  11. anonymous
    • one year ago
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    someone help!

  12. hartnn
    • one year ago
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    write both \(\cos^2 x = \dfrac{1-\cos 2x}{2}\)

  13. anonymous
    • one year ago
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    It must be a typo... I'm getting 1+4cos(2x)+cos(4x) + 3 / 4 sorry.

  14. hartnn
    • one year ago
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    ***\(\cos^2 x = \dfrac{1+\cos 2x}{2}\)

  15. anonymous
    • one year ago
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    elementary mistake hartnn

  16. hartnn
    • one year ago
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    and then expand it.

  17. anonymous
    • one year ago
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    So my attempt leads to something similar to B but with +3 at the end. Either I'm wrong (not likely) or the question is wrong.

  18. hartnn
    • one year ago
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    then using the same identity you'll need to write \(\Large \cos^2 2x = \dfrac{1+\cos 4x}{2}\)

  19. anonymous
    • one year ago
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    ROOKIE MISTAKE!!!

  20. anonymous
    • one year ago
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    sorry to everyone. I am so sorry

  21. anonymous
    • one year ago
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    So I was right

  22. hartnn
    • one year ago
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    so you got it now?

  23. anonymous
    • one year ago
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    i must've made a huge typo

  24. anonymous
    • one year ago
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    the question is 2sin^2 (x) cos^2x

  25. anonymous
    • one year ago
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    sin. not cos. i can't believe i made that mistake.

  26. hartnn
    • one year ago
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    okk, then write \(\large \sin^2 x = \dfrac{1-\cos 2x}{2}\) thats the only difference

  27. anonymous
    • one year ago
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    Dang it I knew you made that mistake! Lol, it gives a difference of two squares at the top. I already did that. The answer is a. Can you see why?

  28. hartnn
    • one year ago
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    I'd like to see sbsbrand's attempt! :)

  29. anonymous
    • one year ago
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    yay it worked! thanks everyone that stayed to help me even tho i wrote the question wrong.......XD

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