anonymous
  • anonymous
the most general solution which satisfies tan a=1, cos a=-1/sqrt()
Mathematics
katieb
  • katieb
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anonymous
  • anonymous
\[\tan \theta=1, \cos \theta=-1/\sqrt{2}\]
anonymous
  • anonymous
\[\theta=\pi/4+\pi+2n \pi={5\pi \over 4}+2n \pi; n=0,1,2,..\]
anonymous
  • anonymous
would you explain it a bit

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anonymous
  • anonymous
First, the angle at which tan is equal 1, and cos is 1/sqrt(2) is pi/4.
anonymous
  • anonymous
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.
anonymous
  • anonymous
So the correspondent angle for pi/4 in the third quadrant is pi/4 +pi.
anonymous
  • anonymous
This angle will occur again after 2n(pi) for n=0,1,2,...
anonymous
  • anonymous
Does that make sense?!
anonymous
  • anonymous
\[2n \pi +3\pi/4\] is it rite?
anonymous
  • anonymous
Yeah, if you have both cos and tan negatives.
anonymous
  • anonymous
nd id i have jus the cos -ve it wud be 5pi/4...is that correct?
anonymous
  • anonymous
*if
anonymous
  • anonymous
Exactly.
anonymous
  • anonymous
kool;-)

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