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anonymous

  • 5 years ago

I doubt anyone knows, but I need to know step by step how to solve the trig equation: Secant [Sine-1 radical2/5] ....please help!

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  1. amistre64
    • 5 years ago
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    the secant of sine inverse (sqrt(2)/5) is the question right? \(sec(sin(sqrt{2}/5)\)

  2. amistre64
    • 5 years ago
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    well: \(sin^{-1}(\sqrt{2}/5)\)

  3. amistre64
    • 5 years ago
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    you should be aware that the sin^-1 takes a number and gives back an angle between 0 and pi

  4. amistre64
    • 5 years ago
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    but easiest thing to do is make a triangle and draw a pic

  5. amistre64
    • 5 years ago
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  6. amistre64
    • 5 years ago
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    the answer is 5 according to the drawing

  7. anonymous
    • 5 years ago
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    my book says 5 radical 23 over 23.... i solve for my triangle taking sine(x)= radical 2 over 5 and find the third side. But from here, i dont know how to find secant

  8. amistre64
    • 5 years ago
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    secant = hyp/adj ; its just the inverse of cosine

  9. amistre64
    • 5 years ago
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    it is this right? \[sec(sin^{-1}(\frac{\sqrt{2}}{5})\]

  10. amistre64
    • 5 years ago
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    i did sqrt(2)^2 = 4 lol ....

  11. anonymous
    • 5 years ago
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    yes, i just dont understand how the answer is 5 radical 23 over 23

  12. amistre64
    • 5 years ago
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    sqrt(25 - 2) = sqrt(23) for the bottom part then right?

  13. anonymous
    • 5 years ago
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    yeah i dont understand how to get to that part. i just got my triangle and then i dont know where to go from there using secant

  14. amistre64
    • 5 years ago
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    if you know 2 sides of a right triangle you can find the thirs by the pythagoreum thrm

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  15. amistre64
    • 5 years ago
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    (sqrt(2))^2 + base^2 = (5)^2 2 + base^2 = 25 base^2 = 25-2 base^2 = 23 base = sqrt(23)

  16. amistre64
    • 5 years ago
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    secant(A) = hyp/base sec = 5/sqrt(23)

  17. anonymous
    • 5 years ago
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    what is base?

  18. anonymous
    • 5 years ago
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    okay i got it, thanks!

  19. amistre64
    • 5 years ago
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    :) youre welcome :)

  20. anonymous
    • 5 years ago
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    I've posted another tricky one if you can solve it for me, that'd be greatly appreciated!

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