What does cos(arctan(-1))+sin(arc csc(-1)) equal?

1 Answer
Mar 10, 2016

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

Explanation:

$\cos \left(\arctan \left(- 1\right)\right)$ means $\cos \alpha$ of an angle $\alpha$, where $\tan \alpha = - 1$. $\tan \alpha = - 1$ for $\alpha = \frac{3 \pi}{4}$ or $\alpha = \frac{- \pi}{4}$.

$\cos \left(\frac{3 \pi}{4}\right) = - \frac{1}{\sqrt{2}}$ and $\cos \left(\frac{- \pi}{4}\right) = \frac{1}{\sqrt{2}}$. Hence $\cos \left(a r c \tan \left(- 1\right)\right) = \pm \frac{1}{\sqrt{2}}$

$\sin \left(a r c \csc \left(- 1\right)\right)$ means $\sin \beta$ of an angle $\beta$, where $\csc \beta = - 1$.

$\csc \beta = - 1$ for $\beta = \frac{3 \pi}{2}$ and $\sin \left(\frac{3 \pi}{2}\right) = - 1$. Hence $\sin \left(a r c \csc \left(- 1\right)\right) = - 1$

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