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anonymous
 5 years ago
the most general solution which satisfies tan a=1, cos a=1/sqrt()
anonymous
 5 years ago
the most general solution which satisfies tan a=1, cos a=1/sqrt()

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\tan \theta=1, \cos \theta=1/\sqrt{2}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\theta=\pi/4+\pi+2n \pi={5\pi \over 4}+2n \pi; n=0,1,2,..\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0would you explain it a bit

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0First, the angle at which tan is equal 1, and cos is 1/sqrt(2) is pi/4.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Then 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
 5 years ago
Best ResponseYou've already chosen the best response.0So the correspondent angle for pi/4 in the third quadrant is pi/4 +pi.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0This angle will occur again after 2n(pi) for n=0,1,2,...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Does that make sense?!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[2n \pi +3\pi/4\] is it rite?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yeah, if you have both cos and tan negatives.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0nd id i have jus the cos ve it wud be 5pi/4...is that correct?
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