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anonymous

  • 5 years ago

the most general solution which satisfies tan a=1, cos a=-1/sqrt()

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  1. anonymous
    • 5 years ago
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    \[\tan \theta=1, \cos \theta=-1/\sqrt{2}\]

  2. anonymous
    • 5 years ago
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    \[\theta=\pi/4+\pi+2n \pi={5\pi \over 4}+2n \pi; n=0,1,2,..\]

  3. anonymous
    • 5 years ago
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    would you explain it a bit

  4. anonymous
    • 5 years ago
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    First, the angle at which tan is equal 1, and cos is 1/sqrt(2) is pi/4.

  5. anonymous
    • 5 years ago
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    Then we should look for the quadrant that has positive values for tan and negative values for cos, and that would be the third quadrant.

  6. anonymous
    • 5 years ago
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    So the correspondent angle for pi/4 in the third quadrant is pi/4 +pi.

  7. anonymous
    • 5 years ago
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    This angle will occur again after 2n(pi) for n=0,1,2,...

  8. anonymous
    • 5 years ago
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    Does that make sense?!

  9. anonymous
    • 5 years ago
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    \[2n \pi +3\pi/4\] is it rite?

  10. anonymous
    • 5 years ago
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    Yeah, if you have both cos and tan negatives.

  11. anonymous
    • 5 years ago
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    nd id i have jus the cos -ve it wud be 5pi/4...is that correct?

  12. anonymous
    • 5 years ago
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    *if

  13. anonymous
    • 5 years ago
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    Exactly.

  14. anonymous
    • 5 years ago
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    kool;-)

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