Difference between revisions of "User:Fath2012"
| Line 6: | Line 6: | ||
HTML test has succeeded! | HTML test has succeeded! | ||
</div> | </div> | ||
| + | <latex> | ||
| + | \begin{align*} &\cos A = \frac{\text{adjacent leg}}{\text{hypotenuse}}, \\ &\sin A = \frac{\text{opposite leg}}{\text{hypotenuse}}, \, \\ &\tan A = \frac{\text{opposite leg}}{\text{adjacent leg}}, \\ &\sec A = \frac{\text{hypotenuse}}{\text{adjacent leg}}, \, \\ &\csc A = \frac{\text{hypotenuse}}{\text{opposite leg}}, \, \\ &\text{ and } \\ &\cot A = \frac{\text{adjacent leg}}{\text{opposite leg}}. \end{align*} | ||
| + | |||
| + | \begin{align*} \sin 30^\circ & = \frac{1}{2}, \\ \cos 30^\circ & = \frac{\sqrt{3}}{2}, \\ \tan 30^\circ & = \frac{\sqrt{3}}{3},\\ \\ \sin 45^\circ & = \frac{\sqrt{2}}{2}, \\ \cos 45^\circ & = \frac{\sqrt{2}}{2}, \\ \tan 45^\circ & = 1, \\ \\ \sin 60^\circ & = \frac{\sqrt{3}}{2}, \\ \cos 60^\circ & = \frac{1}{2}, \\ \text{ and } \\ \tan 60^\circ & = \sqrt{3}. \end{align*} | ||
| + | </latex> | ||
| + | <img src="https://latex.artofproblemsolving.com/8/6/5/865494988ab38b5d8adf1df23b29902b8831d48f.png"></img> | ||
| + | \begin{align*} &\sin^2\theta + \cos^2\theta = 1, \\ &\cos\theta = \cos(-\theta), \\ &\sin\theta = -\sin(-\theta), \\ &\cos\theta = -\cos(180^\circ - \theta), \\ &\text{ and } \\ &\sin\theta = \sin(180^\circ - \theta). \end{align*} | ||
Revision as of 20:09, 21 May 2019
404 not found
Page not found. Maybe the user has deleted it?
HTML test has succeeded!
<latex> \begin{align*} &\cos A = \frac{\text{adjacent leg}}{\text{hypotenuse}}, \\ &\sin A = \frac{\text{opposite leg}}{\text{hypotenuse}}, \, \\ &\tan A = \frac{\text{opposite leg}}{\text{adjacent leg}}, \\ &\sec A = \frac{\text{hypotenuse}}{\text{adjacent leg}}, \, \\ &\csc A = \frac{\text{hypotenuse}}{\text{opposite leg}}, \, \\ &\text{ and } \\ &\cot A = \frac{\text{adjacent leg}}{\text{opposite leg}}. \end{align*}
\begin{align*} \sin 30^\circ & = \frac{1}{2}, \\ \cos 30^\circ & = \frac{\sqrt{3}}{2}, \\ \tan 30^\circ & = \frac{\sqrt{3}}{3},\\ \\ \sin 45^\circ & = \frac{\sqrt{2}}{2}, \\ \cos 45^\circ & = \frac{\sqrt{2}}{2}, \\ \tan 45^\circ & = 1, \\ \\ \sin 60^\circ & = \frac{\sqrt{3}}{2}, \\ \cos 60^\circ & = \frac{1}{2}, \\ \text{ and } \\ \tan 60^\circ & = \sqrt{3}. \end{align*} </latex> <img src="https://latex.artofproblemsolving.com/8/6/5/865494988ab38b5d8adf1df23b29902b8831d48f.png"></img> \begin{align*} &\sin^2\theta + \cos^2\theta = 1, \\ &\cos\theta = \cos(-\theta), \\ &\sin\theta = -\sin(-\theta), \\ &\cos\theta = -\cos(180^\circ - \theta), \\ &\text{ and } \\ &\sin\theta = \sin(180^\circ - \theta). \end{align*}
