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SugarRainbow Group Title

how would you find ^3square root of 38 if you know how I would put that on a calculator that's fine too

  • one year ago
  • one year ago

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  1. miteshchvm Group Title
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    do you have a scienrific calculator?

    • one year ago
  2. sauravshakya Group Title
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    |dw:1347856598297:dw|

    • one year ago
  3. sauravshakya Group Title
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    Type it on calculator

    • one year ago
  4. SugarRainbow Group Title
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    okay please hold

    • one year ago
  5. SugarRainbow Group Title
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    i gots 3.36

    • one year ago
  6. SugarRainbow Group Title
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    just so everyone absolutely knows the problem looks like this |dw:1347856966969:dw|

    • one year ago
  7. sauravshakya Group Title
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    Yes thats it

    • one year ago
  8. SugarRainbow Group Title
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    o kay now how would you find the value of the expression : if sin beta=0.45, find cos(pi/2-beta)

    • one year ago
  9. miteshchvm Group Title
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    its the same

    • one year ago
  10. SugarRainbow Group Title
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    okay then i did it right, what bout if tan(pi/2-beta)=-5.32, find cot beta

    • one year ago
  11. hartnn Group Title
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    tan(pi/2-beta)= cot beta = -5.32

    • one year ago
  12. SugarRainbow Group Title
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    okay cool then I think i got it so if sin (beta-pi/2)=0.73, to find cos-beta i would do cos(sin^-1-pi/2) right? or cos^-1(sin^-1-pi/2)

    • one year ago
  13. hartnn Group Title
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    remember these identities: \(sin(\pi/2-\beta)=cos\beta \\ tan(\pi/2-\beta)=cot\beta \\ cos(\pi/2-\beta)=sin\beta \) when the ratio has angle \(\pi/2 - \beta\) , it becomes coratio.

    • one year ago
  14. SugarRainbow Group Title
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    so is it the same if its beta-pi/2 instead of pi/2-beta?

    • one year ago
  15. hartnn Group Title
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    no, its only for pi/2-beta for beta-pi/2, u write it as -(pi/2-beta) and use sin(-x)=-sin x cos(-x)=cos x tan(-x)=-tan x

    • one year ago
  16. SugarRainbow Group Title
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    okay so to do if sin (beta-pi/2)=0.73, find cos-beta it would be cos(-pi/2-sin^-1(0.73)?

    • one year ago
  17. hartnn Group Title
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    there is no need to do \(sin^{-1}\)here. \(sin(\beta-\pi/2)=sin(-(\pi/2-\beta))=-sin(\pi/2-\beta)=-cos\beta\)

    • one year ago
  18. SugarRainbow Group Title
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    so then how would i do the problem?

    • one year ago
  19. hartnn Group Title
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    so cos beta =- sin (beta-pi/2)= -0.73

    • one year ago
  20. SugarRainbow Group Title
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    oh well IDK how i did it but I got that answer too. okay so last one of these is if cot(-beta)=7.89, find tan( beta-pi/2) would it kinda be like the previous one?

    • one year ago
  21. hartnn Group Title
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    yes. use tan(beta-pi/2) = tan(-(pi/2-beta))=??

    • one year ago
  22. SugarRainbow Group Title
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    please hold

    • one year ago
  23. SugarRainbow Group Title
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    0.03? wait I think i did it wrong let me try again

    • one year ago
  24. SugarRainbow Group Title
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    um okay one last try

    • one year ago
  25. SugarRainbow Group Title
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    1.20?

    • one year ago
  26. hartnn Group Title
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    ok, from where are u getting those numbers ?? tan(beta-pi/2) = tan(-(pi/2-beta))=-tan(pi/2-beta)=-cot beta and cot(-beta)=-cot beta also so tan(beta-pi/2) = cot(-beta) =7.89 see whether u understand this ?

    • one year ago
  27. SugarRainbow Group Title
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    well idk what i put for beta i tried it as just 7.89 and cot(7.89) here i'll try it again

    • one year ago
  28. SugarRainbow Group Title
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    well i understand how it works but idk how to put it into my calc to where i get that answer

    • one year ago
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