6.2 Trigonometric Functions: Unit Circle Approach/14 - Revision history

Dvaezazizi@laalliance.org: Protected "6.2 Trigonometric Functions: Unit Circle Approach/14" ([Edit=Allow only administrators] (indefinite) [Move=Allow only administrators] (indefinite))

2022-08-25T22:19:35Z

Protected "6.2 Trigonometric Functions: Unit Circle Approach/14" ([Edit=Allow only administrators] (indefinite) [Move=Allow only administrators] (indefinite))

← Older revision

Revision as of 22:19, 25 August 2022

2022-08-25T22:19:20Z

← Older revision		Revision as of 22:19, 25 August 2022
Line 1:		Line 1:
	<math>\left(\frac{3}{2}, \frac{\sqrt{1}}{2}\right)</math>		<math>\left(\frac{1}{2}, -\frac{\sqrt{3}}{2}\right)</math>

	<math>		<math>
	\begin{align}		\begin{align}

	\sin{(t)} &= \frac{\sqrt{1}}{2} & \csc{(t)} &= -\frac{2}{\sqrt{3}}\cdot\frac{\sqrt{3}}{\sqrt{3}}=\frac{2\sqrt{3}}{3}\\[2ex]		\sin{(t)} &= -\frac{\sqrt{3}}{2} & \csc{(t)} &= -\frac{2}{\sqrt{3}}\cdot\frac{\sqrt{3}}{\sqrt{3}}=\frac{2\sqrt{3}}{3}\\[2ex]
	\cos{(t)} &= \frac{3}{2} & \sec{(t)} &= \frac{2}{1} = 2\\[2ex]		\cos{(t)} &= \frac{1}{2} & \sec{(t)} &= \frac{2}{1} = 2\\[2ex]
	\tan{(t)} &= {2}}{\frac{3}{2}} = -\frac{\sqrt{3}}{2}\cdot\frac{2}{1} = -\sqrt{3} & \cot{(t)} &= -\frac{1}{\sqrt{3}}=-\frac{1}{\sqrt{3}}\cdot\frac{\sqrt{3}}{\sqrt{3}}=\frac{\sqrt{3}}{3} \\[2ex]		\tan{(t)} &= \frac{-\frac{\sqrt{3}}{2}}{\frac{1}{2}} = -\frac{\sqrt{3}}{2}\cdot\frac{2}{1} = -\sqrt{3} & \cot{(t)} &= -\frac{1}{\sqrt{3}}=-\frac{1}{\sqrt{3}}\cdot\frac{\sqrt{3}}{\sqrt{3}}=\frac{\sqrt{3}}{3} \\[2ex]

	\end{align}		\end{align}
	</math>		</math>
	~~<math>\frac{\frac{\sqrt{1}}<\math>~~

2022-08-25T22:19:08Z

2022-08-25T22:18:13Z

← Older revision		Revision as of 22:18, 25 August 2022
Line 6:		Line 6:
	\sin{(t)} &= \frac{\sqrt{1}}{2} & \csc{(t)} &= -\frac{2}{\sqrt{3}}\cdot\frac{\sqrt{3}}{\sqrt{3}}=\frac{2\sqrt{3}}{3}\\[2ex]		\sin{(t)} &= \frac{\sqrt{1}}{2} & \csc{(t)} &= -\frac{2}{\sqrt{3}}\cdot\frac{\sqrt{3}}{\sqrt{3}}=\frac{2\sqrt{3}}{3}\\[2ex]
	\cos{(t)} &= \frac{3}{2} & \sec{(t)} &= \frac{2}{1} = 2\\[2ex]		\cos{(t)} &= \frac{3}{2} & \sec{(t)} &= \frac{2}{1} = 2\\[2ex]
	\tan{(t)} &= {2}}{\frac{3}{2~~}}\frac{\frac{\sqrt{1~~}} = -\frac{\sqrt{3}}{2}\cdot\frac{2}{1} = -\sqrt{3} & \cot{(t)} &= -\frac{1}{\sqrt{3}}=-\frac{1}{\sqrt{3}}\cdot\frac{\sqrt{3}}{\sqrt{3}}=\frac{\sqrt{3}}{3} \\[2ex]		\tan{(t)} &= {2}}{\frac{3}{2}} = -\frac{\sqrt{3}}{2}\cdot\frac{2}{1} = -\sqrt{3} & \cot{(t)} &= -\frac{1}{\sqrt{3}}=-\frac{1}{\sqrt{3}}\cdot\frac{\sqrt{3}}{\sqrt{3}}=\frac{\sqrt{3}}{3} \\[2ex]

	\end{align}		\end{align}
	</math>		</math>
			\frac{\frac{\sqrt{1}}

2022-08-25T22:15:33Z

← Older revision		Revision as of 22:15, 25 August 2022
Line 6:		Line 6:
	\sin{(t)} &= \frac{\sqrt{1}}{2} & \csc{(t)} &= -\frac{2}{\sqrt{3}}\cdot\frac{\sqrt{3}}{\sqrt{3}}=\frac{2\sqrt{3}}{3}\\[2ex]		\sin{(t)} &= \frac{\sqrt{1}}{2} & \csc{(t)} &= -\frac{2}{\sqrt{3}}\cdot\frac{\sqrt{3}}{\sqrt{3}}=\frac{2\sqrt{3}}{3}\\[2ex]
	\cos{(t)} &= \frac{3}{2} & \sec{(t)} &= \frac{2}{1} = 2\\[2ex]		\cos{(t)} &= \frac{3}{2} & \sec{(t)} &= \frac{2}{1} = 2\\[2ex]
	\tan{(t)} &= ~~\frac~~{\frac{~~\sqrt{1}~~}{2}}{\frac{3}{2}} = -\frac{\sqrt{3}}{2}\cdot\frac{2}{1} = -\sqrt{3} & \cot{(t)} &= -\frac{1}{\sqrt{3}}=-\frac{1}{\sqrt{3}}\cdot\frac{\sqrt{3}}{\sqrt{3}}=\frac{\sqrt{3}}{3} \\[2ex]		\tan{(t)} &= {2}}{\frac{3}{2}}\frac{\frac{\sqrt{1}} = -\frac{\sqrt{3}}{2}\cdot\frac{2}{1} = -\sqrt{3} & \cot{(t)} &= -\frac{1}{\sqrt{3}}=-\frac{1}{\sqrt{3}}\cdot\frac{\sqrt{3}}{\sqrt{3}}=\frac{\sqrt{3}}{3} \\[2ex]

	\end{align}		\end{align}
	</math>		</math>

2022-08-25T22:14:44Z

← Older revision		Revision as of 22:14, 25 August 2022
Line 6:		Line 6:
	\sin{(t)} &= \frac{\sqrt{1}}{2} & \csc{(t)} &= -\frac{2}{\sqrt{3}}\cdot\frac{\sqrt{3}}{\sqrt{3}}=\frac{2\sqrt{3}}{3}\\[2ex]		\sin{(t)} &= \frac{\sqrt{1}}{2} & \csc{(t)} &= -\frac{2}{\sqrt{3}}\cdot\frac{\sqrt{3}}{\sqrt{3}}=\frac{2\sqrt{3}}{3}\\[2ex]
	\cos{(t)} &= \frac{3}{2} & \sec{(t)} &= \frac{2}{1} = 2\\[2ex]		\cos{(t)} &= \frac{3}{2} & \sec{(t)} &= \frac{2}{1} = 2\\[2ex]
	\tan{(t)} &= \frac{-\frac{\sqrt{3}}{2}}{\frac{1}{2}} = -\frac{\sqrt{3}}{2}\cdot\frac{2}{1} = -\sqrt{3} & \cot{(t)} &= -\frac{1}{\sqrt{3}}=-\frac{1}{\sqrt{3}}\cdot\frac{\sqrt{3}}{\sqrt{3}}=\frac{\sqrt{3}}{3} \\[2ex]		\tan{(t)} &= \frac{\frac{\sqrt{1}}{2}}{\frac{3}{2}} = -\frac{\sqrt{3}}{2}\cdot\frac{2}{1} = -\sqrt{3} & \cot{(t)} &= -\frac{1}{\sqrt{3}}=-\frac{1}{\sqrt{3}}\cdot\frac{\sqrt{3}}{\sqrt{3}}=\frac{\sqrt{3}}{3} \\[2ex]

	\end{align}		\end{align}
	</math>		</math>

2022-08-25T22:14:08Z

← Older revision		Revision as of 22:14, 25 August 2022
Line 4:		Line 4:
	\begin{align}		\begin{align}

	\sin{(t)} &= -\frac{\sqrt{1}}{2} & \csc{(t)} &= -\frac{2}{\sqrt{3}}\cdot\frac{\sqrt{3}}{\sqrt{3}}=\frac{2\sqrt{3}}{3}\\[2ex]		\sin{(t)} &= \frac{\sqrt{1}}{2} & \csc{(t)} &= -\frac{2}{\sqrt{3}}\cdot\frac{\sqrt{3}}{\sqrt{3}}=\frac{2\sqrt{3}}{3}\\[2ex]
	\cos{(t)} &= \frac{3}{2} & \sec{(t)} &= \frac{2}{1} = 2\\[2ex]		\cos{(t)} &= \frac{3}{2} & \sec{(t)} &= \frac{2}{1} = 2\\[2ex]
	\tan{(t)} &= \frac{-\frac{\sqrt{3}}{2}}{\frac{1}{2}} = -\frac{\sqrt{3}}{2}\cdot\frac{2}{1} = -\sqrt{3} & \cot{(t)} &= -\frac{1}{\sqrt{3}}=-\frac{1}{\sqrt{3}}\cdot\frac{\sqrt{3}}{\sqrt{3}}=\frac{\sqrt{3}}{3} \\[2ex]		\tan{(t)} &= \frac{-\frac{\sqrt{3}}{2}}{\frac{1}{2}} = -\frac{\sqrt{3}}{2}\cdot\frac{2}{1} = -\sqrt{3} & \cot{(t)} &= -\frac{1}{\sqrt{3}}=-\frac{1}{\sqrt{3}}\cdot\frac{\sqrt{3}}{\sqrt{3}}=\frac{\sqrt{3}}{3} \\[2ex]

2022-08-25T22:13:36Z

← Older revision		Revision as of 22:13, 25 August 2022
Line 1:		Line 1:
	<math>\left(\frac{3}{2}, -\frac{\sqrt{1}}{2}\right)</math>		<math>\left(\frac{3}{2}, \frac{\sqrt{1}}{2}\right)</math>

	<math>		<math>

2022-08-25T22:13:19Z

← Older revision		Revision as of 22:13, 25 August 2022
Line 1:		Line 1:
	<math>\left(\frac{1}{2}, -\frac{\sqrt{3}}{2}\right)</math>		<math>\left(\frac{3}{2}, -\frac{\sqrt{1}}{2}\right)</math>

	<math>		<math>
	\begin{align}		\begin{align}

	\sin{(t)} &= -\frac{\sqrt{3}}{2} & \csc{(t)} &= -\frac{2}{\sqrt{3}}\cdot\frac{\sqrt{3}}{\sqrt{3}}=\frac{2\sqrt{3}}{3}\\[2ex]		\sin{(t)} &= -\frac{\sqrt{1}}{2} & \csc{(t)} &= -\frac{2}{\sqrt{3}}\cdot\frac{\sqrt{3}}{\sqrt{3}}=\frac{2\sqrt{3}}{3}\\[2ex]
	\cos{(t)} &= \frac{1}{2} & \sec{(t)} &= \frac{2}{1} = 2\\[2ex]		\cos{(t)} &= \frac{3}{2} & \sec{(t)} &= \frac{2}{1} = 2\\[2ex]
	\tan{(t)} &= \frac{-\frac{\sqrt{3}}{2}}{\frac{1}{2}} = -\frac{\sqrt{3}}{2}\cdot\frac{2}{1} = -\sqrt{3} & \cot{(t)} &= -\frac{1}{\sqrt{3}}=-\frac{1}{\sqrt{3}}\cdot\frac{\sqrt{3}}{\sqrt{3}}=\frac{\sqrt{3}}{3} \\[2ex]		\tan{(t)} &= \frac{-\frac{\sqrt{3}}{2}}{\frac{1}{2}} = -\frac{\sqrt{3}}{2}\cdot\frac{2}{1} = -\sqrt{3} & \cot{(t)} &= -\frac{1}{\sqrt{3}}=-\frac{1}{\sqrt{3}}\cdot\frac{\sqrt{3}}{\sqrt{3}}=\frac{\sqrt{3}}{3} \\[2ex]

	\end{align}		\end{align}
	</math>		</math>

2022-08-25T22:07:45Z

← Older revision		Revision as of 22:07, 25 August 2022
Line 3:		Line 3:
	<math>		<math>
	\begin{align}		\begin{align}


	\sin{(t)} &= -\frac{\sqrt{3}}{2} & \csc{(t)} &= -\frac{2}{\sqrt{3}}\cdot\frac{\sqrt{3}}{\sqrt{3}}=\frac{2\sqrt{3}}{3}\\[2ex]		\sin{(t)} &= -\frac{\sqrt{3}}{2} & \csc{(t)} &= -\frac{2}{\sqrt{3}}\cdot\frac{\sqrt{3}}{\sqrt{3}}=\frac{2\sqrt{3}}{3}\\[2ex]