Precalc
\sin^2\theta+\cos^2\theta=1
1+\tan^2\theta=\sec^2\theta
\sin2\theta=2\sin\theta\cos\theta
\sin\frac{\pi}{6}=\frac12
\cos\frac{5\pi}{4}=-\frac{1}{\sqrt2}
\tan\frac{5\pi}{6}=-\frac{1}{\sqrt3}
\tan\pi=0
\sin^{-1}\left(-\frac{\sqrt3}{2}\right)=-\frac{\pi}{3}
\cos\left(\frac{3\pi}{2}\right)=0
\cos^{-1}\left(-\frac{1}{\sqrt2}\right)=\frac{3\pi}{4}
\tan^{-1}\left(\tan\frac{5\pi}{4}\right)=\frac{\pi}{4}
\sin^{-1}\left(\sin\frac{11\pi}{6}\right)=-\frac{\pi}{6}
\sin\left(-\frac{3\pi}{2}\right)=1
\cos\pi=-1
An angle of \frac{14\pi}{3} radians is in Quandrant II
An angle of \frac{11\pi}{6} radians is in Quandrant IV
\sec\left(\frac{\pi}{2}\right) is undefined
\sin\frac{5\pi}{4}=-\frac{1}{\sqrt2}
\csc\left(\frac{3\pi}{2}\right)=-1
The period of y=\sin2\theta is \pi
The period of y=\cos\frac13\theta is 6\pi
The period of y=\tan3\theta is \frac{\pi}{3}
\pi radians is 180 degrees
Working in Quandrant I \sin\left(\tan^{-1}\left(\frac{x}{4}\right)\right)=\frac{x}{\sqrt{x^2+16}}