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anonymous

  • 5 years ago

Trig. - Solve for x if 0 < x < 2pi, exact values only in radians. tan ((sin^-1(4/5))/2) please show steps

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  1. anonymous
    • 5 years ago
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    Not sure exactly where your "x" is, but in general: let sin^-1(4/5) = y => sin(y) = 4/5 Use sin^2 + cos^2 = 1 -> 1/tan^2(y) = [1/sin^2(y)] - 1 to find tan(y). Use half angle forumula to find tan(y/2) Blah. can;t be arsed to actually do it though,

  2. anonymous
    • 5 years ago
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    Oppps, it is supposed to be 'Evaluate without a calculator'....Sorry

  3. anonymous
    • 5 years ago
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    I don't see how that effects my method? I don't use calculators, anyway.

  4. anonymous
    • 5 years ago
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    No, it doesn't, it just explains why there is no 'x'. XD

  5. anonymous
    • 5 years ago
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    Thanks!!

  6. anonymous
    • 5 years ago
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    No? I doesn't say what exactly you are solving for. I mean, I assume the whole expression = x, but...

  7. anonymous
    • 5 years ago
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    No it doesn't effect your method is what I meant, and I am solving for tan of (sin^-1(4/5)/2), if that makes any sense. Your method seems right though.

  8. anonymous
    • 5 years ago
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    My Method IS right. It's my method, after all.

  9. anonymous
    • 5 years ago
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    Lol, that it is!

  10. anonymous
    • 5 years ago
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    xD my method is hideously long. Sorry. draw a RA triangle with angle y = arcsin(4/5), you can fill in the other sides {3,4,5} and find sin(y) = 4/5, cos(y) = 3/5 And tan(y/2) = sin(y) / ( 1 + cos (y) ) My bad, haven'y used half tan in a while, but the formulae are all in terms of sin and cos.

  11. anonymous
    • 5 years ago
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    Lol, that works too, thanks soooo much! :)

  12. anonymous
    • 5 years ago
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    You're welcome (the only problem is you may not be able to assume the formula for tan(y/2), but it's not too taxing to prove).

  13. anonymous
    • 5 years ago
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    :)

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