Difference between revisions of "User:Fath2012"
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| − | \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*} &\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*}</math> |
| − | \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*} | + | <math>\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*} |
| − | </ | + | </math> |
<img src="https://latex.artofproblemsolving.com/8/6/5/865494988ab38b5d8adf1df23b29902b8831d48f.png"></img> | <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*} | + | <math>\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*}</math> |
| + | Credits to heatherfinotti for these charts | ||
Revision as of 20:10, 21 May 2019
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$\begin{align*} &\cos A = \frac{\text{adjacent leg}}{\text{hypotenuse}}, \\ &\sin A = \frac{\text{opposite leg}}{\text{hypotenuse}}, \, \\ &\tan A = \frac{\text{opposite leg}}{\text{adjacent leg}}, \\ &\sec A = \frac{\text{hypotenuse}}{\text{adjacent leg}}, \, \\ &\csc A = \frac{\text{hypotenuse}}{\text{opposite leg}}, \, \\ &\text{ and } \\ &\cot A = \frac{\text{adjacent leg}}{\text{opposite leg}}. \end{align*}$ (Error compiling LaTeX. Unknown error_msg)
$\begin{align*} \sin 30^\circ & = \frac{1}{2}, \\ \cos 30^\circ & = \frac{\sqrt{3}}{2}, \\ \tan 30^\circ & = \frac{\sqrt{3}}{3},\\ \\ \sin 45^\circ & = \frac{\sqrt{2}}{2}, \\ \cos 45^\circ & = \frac{\sqrt{2}}{2}, \\ \tan 45^\circ & = 1, \\ \\ \sin 60^\circ & = \frac{\sqrt{3}}{2}, \\ \cos 60^\circ & = \frac{1}{2}, \\ \text{ and } \\ \tan 60^\circ & = \sqrt{3}. \end{align*}$ (Error compiling LaTeX. Unknown error_msg) <img src="https://latex.artofproblemsolving.com/8/6/5/865494988ab38b5d8adf1df23b29902b8831d48f.png"></img> $\begin{align*} &\sin^2\theta + \cos^2\theta = 1, \\ &\cos\theta = \cos(-\theta), \\ &\sin\theta = -\sin(-\theta), \\ &\cos\theta = -\cos(180^\circ - \theta), \\ &\text{ and } \\ &\sin\theta = \sin(180^\circ - \theta). \end{align*}$ (Error compiling LaTeX. Unknown error_msg) Credits to heatherfinotti for these charts
