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anonymous

  • one year ago

What is this TRIG identity called. see attachment.

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  1. anonymous
    • one year ago
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  2. anonymous
    • one year ago
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    @dan815

  3. anonymous
    • one year ago
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    @LynFran

  4. anonymous
    • one year ago
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    @jim_thompson5910

  5. jim_thompson5910
    • one year ago
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    I'm not sure if there is a specific name for this identity

  6. anonymous
    • one year ago
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    how would I explain it though

  7. jim_thompson5910
    • one year ago
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    I guess you could say "composition of a trig function and another inverse" ? I'm not sure

  8. jim_thompson5910
    • one year ago
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    oh you want to know how they got that identity?

  9. anonymous
    • one year ago
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    yeah

  10. anonymous
    • one year ago
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    here let me give you an example

  11. jim_thompson5910
    • one year ago
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    tan(arccos(x)) let theta = arccos(x) so cos(theta) = x = x/1 agreed?

  12. anonymous
    • one year ago
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    \[\tan(\cos^-1(1/2)\]

  13. anonymous
    • one year ago
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    yes agreed

  14. jim_thompson5910
    • one year ago
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    so we can create this right triangle |dw:1438215201249:dw|

  15. jim_thompson5910
    • one year ago
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    |dw:1438215224912:dw|

  16. jim_thompson5910
    • one year ago
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    cos(theta) = x/1 cosine deals with adjacent over hypotenuse

  17. jim_thompson5910
    • one year ago
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    so we can add these labels |dw:1438215259342:dw|

  18. jim_thompson5910
    • one year ago
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    what is the missing side equal to (in terms of x) ?

  19. anonymous
    • one year ago
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    1^2-x^2

  20. jim_thompson5910
    • one year ago
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    it will be the square root of that

  21. jim_thompson5910
    • one year ago
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    |dw:1438215360088:dw|

  22. anonymous
    • one year ago
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    @jim_thompson5910 how do we gate the base of absolute value of x

  23. jim_thompson5910
    • one year ago
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    tan(arccos(x)) turns into tan(theta) this is because I let theta = arccos(x) calculating tan(theta) gives \[\Large \frac{\text{opposite}}{\text{adjacent}} = \frac{\sqrt{1-x^2}}{x}\]

  24. jim_thompson5910
    • one year ago
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    for some reason, and I don't know why, they made the denominator be |x| instead of just x. My guess is that they wanted to force tan(arccos(x)) to be positive. However, tan(arccos(x)) is only positive when x is positive

  25. jim_thompson5910
    • one year ago
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    So it should be \[\Large \tan(\arccos(x)) = \frac{\sqrt{1-x^2}}{x}\]

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