6.2 Trigonometric Functions: Unit Circle Approach/61: Difference between revisions

From Burton Tech. Points Wiki
Jump to navigation Jump to search
No edit summary
No edit summary
 
(2 intermediate revisions by the same user not shown)
Line 1: Line 1:
<math> \frac{5\pi}{6}
<math> \frac{5\pi}{6}
</math>


<math>
\sin({\frac{5\pi}{6}})=1
</math>


\sin{\frac{5\pi}{6}}=1 \sec{\frac{5\pi}{6}}=\frac{1}{1}=1
<math>
\cos{\frac{5\pi}{6}}=0 \csc{\frac{5\pi}{6}}=\frac{1}{0}=Undefined
\sec({\frac{5\pi}{6}})=\frac{1}{1}=1
\tan{\frac{5\pi}{6}}=\frac{1}{0}=Undefind \cot{\frac{5\pi}{6}}=\frac{0}{1}=0
</math>
 
<math>
\cos({\frac{5\pi}{6}})=0  
</math>
 
<math>
\csc({\frac{5\pi}{6}})=\frac{1}{0}=Undefined
</math>
 
<math>
\tan({\frac{5\pi}{6}})=\frac{1}{0}=Undefind  
</math>
 
<math>
\cot({\frac{5\pi}{6}})=\frac{0}{1}=0
</math>
</math>

Latest revision as of 17:22, 6 September 2022