Answer
$\displaystyle \sin(-1470^{o})= -\frac{1}{2}$
$\displaystyle \cos(-1470^{o})= \frac{\sqrt{3}}{2}$
$\displaystyle \tan(-1470^{o})= -\frac{\sqrt{3}}{3}$
$\cot(-1470^{o}) =-\sqrt{3}$
$\displaystyle \sec(-1470^{o})= \frac{2\sqrt{3}}{3}$
$\csc(-1470^{o}) =-2$
Work Step by Step
Reference Angle $\theta^{\prime}$ for $\theta$ in $(0^{\mathrm{o}},\ 360^{\mathrm{o}})$
$\left[\begin{array}{lllll}
Quadrant: & I & II & III & IV\\
\theta' & \theta & 180^{o}-\theta & \theta-180^{o} & 360^{o}-\theta
\end{array}\right]$
$-1470^{o}$ is coterminal with $-1470^{o}+5\cdot 360^{o}$
$=-1470^{o}+1800^{o}=330^{o}$ (quadrant IV. )
$330^{o}$ lies in quadrant IV.
The reference angle is $360^{o}-330^{o}=30^{o}$.
Since $-1470^{o}$ is in quadrant IV,
the cosine and secant are positive,
the sine, tangent, cotangent, and cosecant are negative.
From the Function Values of Special Angles, for $30^{\mathrm{o}}$:
$\displaystyle \sin(-1470^{o})=-\sin 30^{\mathrm{o}}=-\frac{1}{2}$
$\displaystyle \cos(-1470^{o})=\cos 30^{\mathrm{o}}=\frac{\sqrt{3}}{2}$
$\displaystyle \tan(-1470^{o})=-\tan 30^{\mathrm{o}}=-\frac{\sqrt{3}}{3}$
$\cot(-1470^{o})=-\cot 30^{\mathrm{o}}=-\sqrt{3}$
$\displaystyle \sec(-1470^{o})=\sec 30^{\mathrm{o}}=\frac{2\sqrt{3}}{3}$
$\csc(-1470^{o})=-\csc 30^{\mathrm{o}}=-2$
