korosh23
  • korosh23
Pre-cal 12 Question! Trigonometry! radius or r= 3 angle and coordiantes are taking place in quadrant 3 tan theta = 1 Find the missing coordinates!
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
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korosh23
  • korosh23
tan theta = y/ x= 1/1= 1 y = 1 and x= 1 r = 3 |dw:1450243598935:dw| what are the coordinates at point p:
korosh23
  • korosh23
Is my work correct? Is the coordinate (-1, -1 )? Please explain
Directrix
  • Directrix
In quadrant 3, the tangent of an angle is positive. That is because the tangent function is the ratio of sine to cosine and both of those are negative in the third quadrant. -/- = +. Please post the original problem exactly as it appears in your book. Thanks.

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korosh23
  • korosh23
It is a worksheet, and believe me this is what is on the paper: radius or r= 3 angle and coordiantes are taking place in quadrant 3 tan theta = 1 Find the missing coordinates!
johnweldon1993
  • johnweldon1993
Well, \(\large tan(x) = \frac{sin(x)}{cos(x)}\) since we know tan(x) = 1...then we need the ratio of sin to cos to equal 1...or...we need to know when sin(x) = cos(x) which coincidentally occur at an angle of 45 degrees However, the problem is 45 degrees in in the first quadrant...so to move to the third quadrant we add 180 degrees and make it 225 degrees Now, with that angle known, what would be the coordinates?
korosh23
  • korosh23
In this situation the coordinates are \[(r \sin \theta, r \cos \theta )\] \[\sin \theta = - \sqrt{2}/2\] \[\cos \theta = - \sqrt{2}/2\] then times by 3 which results in (-3squareroot2/2, -3 squareroot 2/2)
korosh23
  • korosh23
I also found a new method x^2 + y^2 = r^2 since x = y in this question 2x^2= 3^2 x= square root 9/2 = 3/squareroot 2 = 3.squareroot2/2 This looks confusing. sorry :D

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