What does $sin (a r c cos (2)) + 3 cos (arctan (- 1))$ $sin(arc cos(2))+3cos(arctan(-1))$ equal?

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Answer 1 · 2015-12-19T14:10:18

Nothing.

$arccos$ $arccos$ is a function that is only defined on $[- 1, 1]$ $[-1,1]$ so $arccos (2)$ $arccos(2)$ doesn't exist.

On the other hand, $arctan$ $arctan$ is defined on $RR$ so $arctan(-1)$ exists. It is an odd function so $arctan(-1) = -arctan(1) = -pi/4$ .

So $3cos(arctan(-1)) = 3cos(-pi/4) = 3cos(pi/4) = (3sqrt(2))/2$ .

What does sin(arc cos(2))+3cos(arctan(-1))sin(arccos(2))+3cos(arctan(−1))sin(arc cos(2))+3cos(arctan(-1)) equal?