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cutiecomittee123

  • one year ago

What functions are equivelent to y=4cosx-2 y=4cos(-x)-2 y=4sin(x+pi/2)-2 y=4sin(x-pi/2)-2 y=-4cosx+2

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  1. jango_IN_DTOWN
    • one year ago
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    cos(-x)=cos x . now try

  2. imqwerty
    • one year ago
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    these are some identities try using them to get the answer \[\cos(-x)=\cos(x)\]\[\sin \left( \frac{ \pi }{ 2 }+x \right)=\sin(x)\]\[\sin \left( \frac{ \pi }{ 2 }-x \right)=\cos(x)\]

  3. cutiecomittee123
    • one year ago
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    so a and d work in that case?

  4. imqwerty
    • one year ago
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    d won't work

  5. cutiecomittee123
    • one year ago
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    how so?

  6. cutiecomittee123
    • one year ago
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    is it because its not the x that is being negative, its the whole equation?

  7. cutiecomittee123
    • one year ago
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    and is that the only one that is equal?

  8. imqwerty
    • one year ago
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    so if d is the answer -4cos(x)+2=4cos(x)-2 bring both on right hand side 0=8cos(x)-4 is this correct no its not so d won't wrk what do u think about c

  9. cutiecomittee123
    • one year ago
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    oh i see, so if you just set them equal to eachother and solve then you will get your answers!!!!!

  10. imqwerty
    • one year ago
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    yea u can check the options like that

  11. imqwerty
    • one year ago
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    if the option u choose satisfies the equation then its correct :D

  12. cutiecomittee123
    • one year ago
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    cool thanks

  13. jim_thompson5910
    • one year ago
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    You can also use a graphing calculator to check. https://www.desmos.com/calculator/lwlzconupi Notice how the original graph of `y=4cos(x)-2` (in red) does not match with choice D `y=-4cos(x)+2` (in blue). If they did match, then one curve would be right on top of the other. You can turn on/off the graphs to see if one was on top of the other.

  14. jim_thompson5910
    • one year ago
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    this shows how the original graph matches up with choice A https://www.desmos.com/calculator/gwwpecq6te click the circle icons to turn the graph on/off and you'll see one graph overlapping perfectly on the other

  15. cutiecomittee123
    • one year ago
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    well b and c cannot match because they are sin and the origional is cosine, right?

  16. cutiecomittee123
    • one year ago
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    actually i just tested it and b actually matches up

  17. jim_thompson5910
    • one year ago
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    btw @imqwerty it should be sin(pi/2 + x) = cos(x)

  18. cutiecomittee123
    • one year ago
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    https://www.desmos.com/calculator/lwlzconupi

  19. jim_thompson5910
    • one year ago
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    yeah B should match with the original

  20. cutiecomittee123
    • one year ago
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    so b and d are both equal

  21. imqwerty
    • one year ago
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    >.< oh srry silly mistake

  22. jim_thompson5910
    • one year ago
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    `so b and d are both equal` incorrect

  23. cutiecomittee123
    • one year ago
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    yeah but b and d match up

  24. cutiecomittee123
    • one year ago
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    so then just d? was I right about the fact that they cannot be the same because b and c are sine and the origional is cosine?

  25. jim_thompson5910
    • one year ago
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    since sin(pi/2 + x) = cos(x), this means y=4sin(x+pi/2)-2 turns into y=4cos(x)-2

  26. cutiecomittee123
    • one year ago
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    okay i kind of get that so does that mean that c works

  27. jim_thompson5910
    • one year ago
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    one moment

  28. cutiecomittee123
    • one year ago
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    i graphed it and they matched up

  29. jim_thompson5910
    • one year ago
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    ok I graphed the original equation (in box 1) and the four answer choices (boxes 2 through 5). Only box 1 is turned on. The other graphs are turned off https://www.desmos.com/calculator/koejh8yigm I recommend going through each and turn them on one at a time. example: turn on box 1 and box 3 to compare the original with choice B

  30. jim_thompson5910
    • one year ago
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    hopefully you'll be able to see that A matches B matches C does not match D does not match

  31. cutiecomittee123
    • one year ago
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    yep that what i see too. I tested it out

  32. jim_thompson5910
    • one year ago
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    ok great

  33. cutiecomittee123
    • one year ago
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    sweet thanks

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