Precalc

\sin^2\theta+\cos^2\theta=

1

1+\tan^2\theta=

\sec^2\theta

\sec^2\theta =

1+\tan^2\theta

\sin2\theta =

2\sin\theta\cos\theta

2\sin\theta\cos\theta =

\sin2\theta

\cos2\theta =

\cos^2\theta-\sin^2\theta

\cos^2\theta-\sin^2\theta =

\cos2\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}

\sec\frac{4\pi}{3} =

-2

\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 which Quandrant

Quandrant II

An angle of \frac{11\pi}{6} radians is in which Quandrant

Quandrant IV

\sec\left(\frac{\pi}{2}\right) =

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 how many degrees

180 degrees

180 degrees is how many radians

\pi radians

Working in Quadrant I, \sin\left(\tan^{-1}\left(\frac{x}{4}\right)\right) =

\frac{x}{\sqrt{x^2+16}}