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, $\sin\left(\tan^{-1}\left(\frac{x}{4}\right)\right)$ $=$

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