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anonymous

  • one year ago

Find the exact value of the inverse trigonometric function cos(sin^-1(root 2/3))

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  1. anonymous
    • one year ago
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    @jim_thompson5910

  2. jim_thompson5910
    • one year ago
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    Let \[\Large \theta = \sin^{-1}\left(\frac{\sqrt{2}}{3}\right)\] So that means \[\Large \sin(\theta) = \frac{\sqrt{2}}{3}\]

  3. jim_thompson5910
    • one year ago
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    since \[\Large \sin(\theta) = \frac{\sqrt{2}}{3}\] this means we can draw out this triangle |dw:1433388810567:dw|

  4. jim_thompson5910
    • one year ago
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    what is the missing side?

  5. anonymous
    • one year ago
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    wait how did you know what value corresponds with what angle/side?

  6. jim_thompson5910
    • one year ago
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    because sine = opposite/hypotenuse |dw:1433388989918:dw|

  7. anonymous
    • one year ago
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    oh ok so you wouldn't adjacent/hypotenuse cause there's a cos?

  8. jim_thompson5910
    • one year ago
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    well that comes later into play, but when setting up sine, you only deal with opposite/hypotenuse

  9. anonymous
    • one year ago
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    okay, sorry I might ask a lot of question as a forewarning

  10. jim_thompson5910
    • one year ago
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    that's fine

  11. jim_thompson5910
    • one year ago
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    what's the missing side?

  12. anonymous
    • one year ago
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    root 7?

  13. jim_thompson5910
    • one year ago
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    yep

  14. jim_thompson5910
    • one year ago
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    |dw:1433389408003:dw|

  15. jim_thompson5910
    • one year ago
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    so saying \[\Large \cos\left(\sin^{-1}\left(\frac{\sqrt{2}}{3}\right)\right)\] is the same exact thing as saying \[\Large \cos\left(\theta\right)\]

  16. jim_thompson5910
    • one year ago
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    Because remember I said "Let \[\Large \theta = \sin^{-1}\left(\frac{\sqrt{2}}{3}\right)\]" back up at the top

  17. anonymous
    • one year ago
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    yeah

  18. jim_thompson5910
    • one year ago
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    so whats the final answer

  19. anonymous
    • one year ago
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    so is it just root 7? or am I misinterpreting?

  20. jim_thompson5910
    • one year ago
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    close

  21. jim_thompson5910
    • one year ago
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    |dw:1433389714910:dw|

  22. anonymous
    • one year ago
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    OHHHH ok I get it now~ so it would be root 7/3?

  23. jim_thompson5910
    • one year ago
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    yep \[\Large \frac{\sqrt{7}}{3}\]

  24. anonymous
    • one year ago
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    cool! Thankyou!

  25. jim_thompson5910
    • one year ago
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    yw

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