Answer
$\displaystyle \sin(-225^{\mathrm{o}})= =\frac{\sqrt{2}}{2}$
$\displaystyle \cos(-225^{\mathrm{o}})= -\frac{\sqrt{2}}{2}$
$\tan(-225^{\mathrm{o}})= -1$
$\cot(-225^{\mathrm{o}})= -1$
$\sec(-225^{\mathrm{o}})= -\sqrt{2}$
$\csc(-225^{\mathrm{o}}) =\sqrt{2}$
Work Step by Step
Reference Angle $\theta^{\prime}$ for $\theta$ in $(0^{\mathrm{o}},\ 360^{\mathrm{o}})$
$\left[\begin{array}{lllll}
Quadr.: & I & II & III & IV\\
\theta' & \theta & 180^{o}-\theta & \theta-180^{o} & 360^{o}-\theta
\end{array}\right]$
$-225^{0}$ is coterminal with $-225^{0}+360^{0}=135^{\mathrm{o}}$.
$135^{\mathrm{o}}$ lies in quadrant II.
The reference angle is$ 180^{o}-135^{\mathrm{o}}=45^{\mathrm{o}}$.
Since $-225^{\mathrm{o}}$ is in quadrant II,
the sine and cosecant are positive,
the cosine, tangent, cotangent, and secant are negative.
From the Function Values of Special Angles, for $45^{\mathrm{o}}$:
$\displaystyle \sin(-225^{\mathrm{o}})=\sin 45^{\mathrm{o}}=\frac{\sqrt{2}}{2}$
$\displaystyle \cos(-225^{\mathrm{o}})=-\cos 45^{\mathrm{o}}=-\frac{\sqrt{2}}{2}$
$\tan(-225^{\mathrm{o}})=-\tan 45^{\mathrm{o}}=-1$
$\cot(-225^{\mathrm{o}})=-\cot 45^{\mathrm{o}}=-1$
$\sec(-225^{\mathrm{o}})=-\sec 45^{\mathrm{o}}=-\sqrt{2}$
$\csc(-225^{\mathrm{o}})=\csc 45^{\mathrm{o}}=\sqrt{2}$
