How do you find the values of the other trigonometric functions of θ from the information given #cos theta = − 5/13# , and theta is in quadrant III?

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Answer 1 · 2015-05-30T20:00:45

This comes from a #5#, #12#, #13# triangle (#5^2+12^2 = 13^2#)

#sin theta = -12/13# (#sin < 0# in QIII)

#tan theta = sin theta / cos theta = 5/12#

#cot theta = cos theta / sin theta = 12/5#

#sec theta = 1 / cos theta = -13/5#

#csc theta = 1 / sin theta = -13/12#

Incidentally, the #5-12-13# triangle is the next right angled triangle in a sequence that begins with the #3-4-5# triangle.

#a_k = 2k+3#

#b_k = (a_k^2-1)/2 = 2k^2+6k+4#

#c_k = (a_k^2+1)/2 = 2k^2+6k+5#