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

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

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Answer 1 · 2016-03-10T08:55:05

$cos (arctan (- 1)) + sin (a r c csc (- 1)) = - 1 \pm \frac{1}{\sqrt{2}}$ $cos(arctan(-1))+sin(arc csc(-1))=-1+-1/sqrt2$

Explanation:

$cos (arctan (- 1))$ $cos(arctan(-1))$ means $cos α$ $cosalpha$ of an angle $α$ $alpha$ , where $tan α = - 1$ $tanalpha=-1$ . $tan α = - 1$ $tanalpha=-1$ for $α = \frac{3 π}{4}$ $alpha=(3pi)/4$ or $α = \frac{- π}{4}$ $alpha=(-pi)/4$ .

$cos (\frac{3 π}{4}) = - \frac{1}{\sqrt{2}}$ $cos((3pi)/4)=-1/sqrt2$ and $cos (\frac{- π}{4}) = \frac{1}{\sqrt{2}}$ $cos((-pi)/4)=1/sqrt2$ . Hence $cos (a r c tan (- 1)) = \pm \frac{1}{\sqrt{2}}$ $cos(arc tan(-1))=+-1/sqrt2$

$sin (a r c csc (- 1))$ $sin(arc csc(-1))$ means $sin β$ $sinbeta$ of an angle $β$ $beta$ , where $csc β = - 1$ $cscbeta=-1$ .

$csc β = - 1$ $cscbeta=-1$ for $β = \frac{3 π}{2}$ $beta=(3pi)/2$ and $sin (\frac{3 π}{2}) = - 1$ $sin((3pi)/2)=-1$ . Hence $sin (a r c csc (- 1)) = - 1$ $sin(arc csc(-1))=-1$

Hence, $cos (arctan (- 1)) + sin (a r c csc (- 1)) = - 1 \pm \frac{1}{\sqrt{2}}$ $cos(arctan(-1))+sin(arc csc(-1))=-1+-1/sqrt2$

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