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anonymous

  • one year ago

Thanks for helping! 6. Prove the identity: (cosx+cosy)^2+(sinx-siny)^2=2+2cos(x+y) Part I: Complete the left-hand column of the table below following the steps indicated in the right-hand column. (12 points) Given on the left side of the original problem (cosx+cosy)^2+(sinx-siny)^2 Expand each squared term Simplify the expression Apply the Pythagorean identity Apply the addition formula for cosine

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  1. anonymous
    • one year ago
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    @dan815 @Kainui @jim_thompson5910 @zepdrix

  2. dan815
    • one year ago
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    expand left side

  3. dan815
    • one year ago
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    (cosx+cosy)^2+(sinx-siny)^2=??

  4. anonymous
    • one year ago
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    2+2cos(x+y)

  5. dan815
    • one year ago
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    no i mean expand it

  6. anonymous
    • one year ago
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    oh gotcha

  7. dan815
    • one year ago
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    and use the cos trig identities

  8. dan815
    • one year ago
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    cosxcosy+sinxsiny = cos(x+y) ?

  9. anonymous
    • one year ago
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    cos^2(x)+2cos(x)cos(y)-2sin(x)sin(x)+cos^2(x)+sin^2(y)+sin^2

  10. anonymous
    • one year ago
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    That is the identity?

  11. dan815
    • one year ago
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    i dunno maybe its a negative inbetween

  12. dan815
    • one year ago
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    cosxcosy- sinxsiny = cos(x+y) ?

  13. dan815
    • one year ago
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    http://prntscr.com/7daa8n

  14. dan815
    • one year ago
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    check it out

  15. anonymous
    • one year ago
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    thanks, from here I can simplify and use the Pythagorean identity?

  16. dan815
    • one year ago
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    yep

  17. dan815
    • one year ago
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    sin^2(a) + cos^2(a) = 1;

  18. dan815
    • one year ago
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    |dw:1433472978378:dw|

  19. dan815
    • one year ago
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    as seen from the unit circle

  20. anonymous
    • one year ago
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    thanks!

  21. anonymous
    • one year ago
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    I'll give you a medal. That helped a lot!

  22. anonymous
    • one year ago
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    I'm new to this site, so I think that is what I'm supposed to do. Ha ha

  23. dan815
    • one year ago
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    okay :)

  24. dan815
    • one year ago
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    you're welcome

  25. anonymous
    • one year ago
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    :D

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