anonymous
  • anonymous
Find the exact value of each expression: a) tan^(-1) (1/sqrt3) b) sec^(-1) (2)
Mathematics
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anonymous
  • anonymous
Find the exact value of each expression: a) tan^(-1) (1/sqrt3) b) sec^(-1) (2)
Mathematics
katieb
  • katieb
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anonymous
  • anonymous
best cheat sheet is here: http://tutorial.math.lamar.edu/cheat_table.aspx
anonymous
  • anonymous
first one you are looking for a number (angle) between \[-\frac{\pi}{2}\] and \[\frac{\pi}{2}\]whose tangent is \[\frac{1}{\sqrt{3}}\]
anonymous
  • anonymous
from the cheat sheet we see that at \[\frac{\pi}{6}\] we get \[sin(\frac{\pi}{6})=\frac{1}{2}\] and \[cos(\frac{\pi}{6})=\frac{\sqrt{3}}{2}\] and therefore \[tan(\frac{\pi}{6})=\frac{1}{\sqrt{3}}\]

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anonymous
  • anonymous
so \[tan^{-1}(\frac{1}{\sqrt{3}})=\frac{\pi}{6}\]
anonymous
  • anonymous
of course you could just use a calculator. are you working in degrees or radian?
anonymous
  • anonymous
anyway second one is easier. \[sec^{-1}(2)\] is the number whose cosine is \[\frac{1}{2}\] and that is \[\frac{\pi}{3}\]

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