6.2 Trigonometric Functions: Unit Circle Approach/41: Difference between revisions

Latest revision as of 17:59, 6 September 2022

$2(\sin({\frac {\pi }{3}}))-3(\tan({\frac {\pi }{6}}))$

$2({\frac {\sqrt {3}}{2}})-3({\frac {\frac {1}{\cancel {2}}}{\frac {\sqrt {3}}{\cancel {2}}}})$

$2({\frac {\sqrt {3}}{2}})-3({\frac {1}{\sqrt {3}}})$

${\frac {2{\sqrt {3}}}{2}}-{\frac {3}{\sqrt {3}}}({\sqrt {3}})$

${\frac {{\cancel {2}}{\sqrt {3}}}{\cancel {2}}}-{\frac {{\cancel {3}}{\sqrt {3}}}{\cancel {3}}}$

${\sqrt {3}}-{\sqrt {3}}=0$

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-hi<br><math>2sin{\frac{{\pi}{3}}<\math>
+<math>2(\sin(\frac{\pi}{3}))-3(\tan(\frac{\pi}{6}))</math>
+<math>2(\frac{\sqrt{3}}{2})-3(\frac{\frac{1}{\cancel{2}}}{\frac{\sqrt{3}}{\cancel{2}}})
+</math>
+<math>
+(\frac{\sqrt{3}}{2})-3(\frac{1}{\sqrt{3}})
+</math>
+<math>
+\frac{2\sqrt{3}}{2}-\frac{3}{\sqrt{3}}(\sqrt{3})
+</math>
+<math>
+\frac{\cancel{2}\sqrt{3}}{\cancel{2}}-\frac{\cancel{3}\sqrt{3}}{\cancel{3}}
+</math>
+<math>
+\sqrt{3}-\sqrt{3}=0
+</math>