Answer
$\displaystyle \sin(-135^{o})=-\frac{\sqrt{2}}{2}$
$\displaystyle \cos(-135^{o})=-\frac{\sqrt{2}}{2}$
$\tan(-135^{o})=1$
$\csc(-135^{o})=-\sqrt{2}$
$\sec(-135^{o})=-\sqrt{2}$
$\cot(-135^{o})=1$
Work Step by Step
Reference Angle $\theta^{\prime}$ for $\theta$ in $(0^{\mathrm{o}},\ 360^{\mathrm{o}})$
$\left[\begin{array}{lllll}
Quadrant: & I & II & III & IV\\
\theta' & \theta & 180^{o}-\theta & \theta-180^{o} & 0^{o}-\theta
\end{array}\right]$
$-135^{o}$ is between $-90^{o}$ and $-180^{o}$ ,in quadrant III,
and has the same terminal side as the angle
$-135^{o}+360^{o}=225^{o}$
for which the reference angle is
$\theta^{\prime}=\theta-180^{o}=225^{o}-180^{o}=45^{o}$
Signs:
In quadrant III, tan and cot are positive, all the others are negative
We use the table Function Values of Special Angles with the corresponding signs:
$\displaystyle \sin(-135^{o})=-\sin 45^{o}=-\frac{\sqrt{2}}{2}$
$\displaystyle \cos(-135^{o})=-\cos 45^{o}=-\frac{\sqrt{2}}{2}$
$\tan(-135^{o})=\tan 45^{o}=1$
$\csc(-135^{o})=-\csc 45^{o}=-\sqrt{2}$
$\sec(-135^{o})=-\sec 45^{o}=-\sqrt{2}$
$\cot(-135^{o})=\cot 45^{o}=1$
