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haleyelizabeth2017

  • one year ago

Trig :)

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  1. haleyelizabeth2017
    • one year ago
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  2. anonymous
    • one year ago
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    Coz f(x) have one half of the period of g(x) , f(x) have a period of pi. and f(pi/4) = 4 and they are both sine waves. f(0)=0 and because of period , x=pi f(pi) = 0 again. so f(x) = 4sin(2x)

  3. anonymous
    • one year ago
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    we can assume f(x) =ASin(Bx) +C where A,B,C are variables x=0, f(x)=0 and when x=pi/4 f(x)=4 and when x= pi f(x) =0 so we can solve it for the three variables that way too

  4. haleyelizabeth2017
    • one year ago
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    I'm confused...

  5. anonymous
    • one year ago
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    sorry i assigned f(x) as f(θ ) and g(x) as g(θ )

  6. haleyelizabeth2017
    • one year ago
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    It's not that....

  7. haleyelizabeth2017
    • one year ago
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    I'm just not quite understanding what I have to do :(

  8. anonymous
    • one year ago
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    ok lets try from the beginning. It says both functions are sine functions. right? so f(θ ) takes the format of f(θ )= ASin(Bx)+C

  9. haleyelizabeth2017
    • one year ago
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    I've never really learned that format....I just learned that "A" as you say is the amplitude, and B is period...what is C? O.o

  10. haleyelizabeth2017
    • one year ago
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    Wait...that might just be for cos. Never mind...

  11. anonymous
    • one year ago
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    So can you solve it now?

  12. haleyelizabeth2017
    • one year ago
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    Lemme try.... :)

  13. haleyelizabeth2017
    • one year ago
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    I'm just caught on the amplitude part....

  14. haleyelizabeth2017
    • one year ago
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    Which one is amp.... pi/4 or 4?

  15. haleyelizabeth2017
    • one year ago
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    I think 4....

  16. anonymous
    • one year ago
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    yeah amplitude is 4

  17. haleyelizabeth2017
    • one year ago
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    \(g(\theta)=2~sin~2\pi~\theta\) eh?

  18. anonymous
    • one year ago
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    just relax its easy . this is a graph ok? a graph of a sin function. according to θ ,y or f(x) change ok?

  19. haleyelizabeth2017
    • one year ago
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    Ah, yes.

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