Question 970477: Evaluate cos[arctan(-5/12)].
Found 3 solutions by lwsshak3, ikleyn, n2:
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! Evaluate cos[arctan(-5/12)]. Interval [0, π]
let x=angle whose tan is (-5/12)
tanx=(-5/12) (working with a (5-12-13) right triangle in quadrant II)
cos[arctan(-5/12)]=cosx=-12/13
Answer by ikleyn(53875)  (Show Source):
You can put this solution on YOUR website! .
Evaluate cos[arctan(-5/12)].
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In the post by @lwsshak3 the answer is incorrect and reasoning is also incorrect.
See below my correct solution.
Function arctan has values in the interval ( , ).
Therefore, the problem asks to evaluate cos(x), given that 'x' is in the fourth quadrant
(not in the second quadrant, as @lwsshak3 mistakenly assumes) and tan(x) = -5/12.
So, let x=angle in QIV whose tan is (-5/12)
tan(x) = (-5/12) (working with a (5-12-13) right triangle in quadrant IV)
cosine is positive in QIV; therefore
cos[arctan(-5/12)] = cos(x) = 12/13.
ANSWER. cos[arctan(-5/12)] = 12/13.
Solved correctly.
Answer by n2(88) (Show Source):
You can put this solution on YOUR website! .
Evaluate cos[arctan(-5/12)].
~~~~~~~~~~~~~~~~~~~~~~~~~~
Function arctan has values in the interval (,).
Therefore, the problem asks to evaluate cos(x), given that 'x' is in the fourth quadrant
and tan(x) = -5/12.
So, let x=angle in QIV whose tan is (-5/12)
tan(x) = (-5/12) (working with a (5-12-13) right triangle in quadrant IV)
cosine is positive in QIV; therefore
cos[arctan(-5/12)] = cos(x) = 12/13.
ANSWER. cos[arctan(-5/12)] = 12/13.
Solved.
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