6.2 Trigonometric Functions: Unit Circle Approach/79

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Revision as of 15:49, 7 September 2022 by Kevind75816@students.laalliance.org (talk | contribs) (Created page with " <math> \theta \rightarrow x=-3, \, y=4, \, r=5 </math><br><br> <math> -3^2 + 4^2 = 5^2 </math><br><br> <math> 9 + 16 = 25 </math><br><br> <math>\sqrt{25} = 5 = r </math><br><br> <math> \begin{align} \sin{(\theta)} &= \frac{4}{5} & \csc{(\theta)} &= \frac{5}{4}\\[2ex] \cos{(\theta)} &= \frac{-3}{5} & \sec{(\theta)} &= \frac{5}{-3}\\[2ex] \tan{(\theta)} &= \frac{4}{-3} & \cot{(\theta)} &= \frac{-3}{4} \\[2ex] \end{align} </math>")
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Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle \theta \rightarrow x=-3, \, y=4, \, r=5 } 

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle  -3^2 + 4^2 = 5^2 }

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle 9 + 16 = 25 }

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle \sqrt{25} = 5 = r }

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle \begin{align} \sin{(\theta)} &= \frac{4}{5} & \csc{(\theta)} &= \frac{5}{4}\\[2ex] \cos{(\theta)} &= \frac{-3}{5} & \sec{(\theta)} &= \frac{5}{-3}\\[2ex] \tan{(\theta)} &= \frac{4}{-3} & \cot{(\theta)} &= \frac{-3}{4} \\[2ex] \end{align} }

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