In Exercises
exactvaluearcfirst -
exactvaluearclast , find the exact value. (To enter
\(\pi \) in an answer box, type “pi”.)
\(\arcsin \left ( -1 \right )\)
\(\arcsin \left ( -1 \right ) = -\frac {\pi }{2}\)
\(\arcsin \left ( -\frac {\sqrt {3}}{2} \right )\)
\(\arcsin \left ( -\frac {\sqrt {3}}{2} \right ) = \answer {-\frac {\pi }{3}}\)
\(\arcsin \left ( -\frac {\sqrt {2}}{2} \right )\)
\(\arcsin \left ( -\frac {\sqrt {2}}{2} \right ) = -\frac {\pi }{4}\)
\(\arcsin \left ( -\frac {1}{2} \right )\)
\(\arcsin \left ( -\frac {1}{2} \right ) = \answer {-\frac {\pi }{6}}\)
\(\arcsin \left ( 0 \right )\)
\(\arcsin \left ( 0 \right ) = 0\)
\(\arcsin \left ( \frac {1}{2} \right )\)
\(\arcsin \left ( \frac {1}{2} \right ) = \answer {\frac {\pi }{6}}\)
\(\arcsin \left ( \frac {\sqrt {2}}{2} \right )\)
\(\arcsin \left ( \frac {\sqrt {2}}{2} \right ) = \frac {\pi }{4}\)
\(\arcsin \left ( \frac {\sqrt {3}}{2} \right )\)
\(\arcsin \left ( \frac {\sqrt {3}}{2} \right ) = \answer {\frac {\pi }{3}}\)
\(\arcsin \left ( 1 \right )\)
\(\arcsin \left ( 1 \right ) = \frac {\pi }{2}\)
\(\arccos \left ( -1 \right )\)
\(\arccos \left ( -1 \right ) = \answer {\pi }\)
\(\arccos \left ( -\frac {\sqrt {3}}{2} \right )\)
\(\arccos \left ( -\frac {\sqrt {3}}{2} \right ) = \frac {5\pi }{6}\)
\(\arccos \left ( -\frac {\sqrt {2}}{2} \right )\)
\(\arccos \left ( -\frac {\sqrt {2}}{2} \right ) = \answer {\frac {3\pi }{4}}\)
\(\arccos \left ( -\frac {1}{2} \right )\)
\(\arccos \left ( -\frac {1}{2} \right ) = \frac {2\pi }{3}\)
\(\arccos \left ( 0 \right )\)
\(\arccos \left ( 0 \right ) = \answer {\frac {\pi }{2}}\)
\(\arccos \left ( \frac {1}{2} \right )\)
\(\arccos \left ( \frac {1}{2} \right ) = \frac {\pi }{3}\)
\(\arccos \left ( \frac {\sqrt {2}}{2} \right )\)
\(\arccos \left ( \frac {\sqrt {2}}{2} \right ) = \answer {\frac {\pi }{4}}\)
\(\arccos \left ( \frac {\sqrt {3}}{2} \right )\)
\(\arccos \left ( \frac {\sqrt {3}}{2} \right ) = \frac {\pi }{6}\)
\(\arccos \left ( 1 \right )\)
\(\arccos \left ( 1 \right ) = \answer {0}\)
\(\arctan \left ( -\sqrt {3} \right )\)
\(\arctan \left ( -\sqrt {3} \right ) = -\frac {\pi }{3}\)
\(\arctan \left ( -1 \right )\)
\(\arctan \left ( -1 \right ) = \answer {-\frac {\pi }{4}}\)
\(\arctan \left ( -\frac {\sqrt {3}}{3} \right )\)
\(\arctan \left ( -\frac {\sqrt {3}}{3} \right ) = -\frac {\pi }{6}\)
\(\arctan \left ( 0 \right )\)
\(\arctan \left ( 0 \right ) = \answer {0}\)
\(\arctan \left ( \frac {\sqrt {3}}{3} \right )\)
\(\arctan \left ( \frac {\sqrt {3}}{3} \right ) = \frac {\pi }{6}\)
\(\arctan \left ( 1 \right )\)
\(\arctan \left ( 1 \right ) = \answer {\frac {\pi }{4}}\)
\(\arctan \left ( \sqrt {3} \right )\)
\(\arctan \left ( \sqrt {3} \right ) = \frac {\pi }{3}\)
\(\mbox {arccot} \left ( -\sqrt {3} \right )\)
\(\mbox {arccot} \left ( -\sqrt {3} \right ) = \answer {\frac {5\pi }{6}}\)
\(\mbox {arccot} \left ( -1 \right )\)
\(\mbox {arccot} \left ( -1 \right ) = \frac {3\pi }{4}\)
\(\mbox {arccot} \left ( -\frac {\sqrt {3}}{3} \right )\)
\(\mbox {arccot} \left ( -\frac {\sqrt {3}}{3} \right ) = \answer {\frac {2\pi }{3}}\)
\(\mbox {arccot} \left ( 0 \right )\)
\(\mbox {arccot} \left ( 0 \right ) = \frac {\pi }{2}\)
\(\mbox {arccot} \left ( \frac {\sqrt {3}}{3} \right )\)
\(\mbox {arccot} \left ( \frac {\sqrt {3}}{3} \right ) = \answer {\frac {\pi }{3}}\)
\(\mbox {arccot} \left ( 1 \right )\)
\(\mbox {arccot} \left ( 1 \right ) = \frac {\pi }{4}\)
\(\mbox {arccot} \left ( \sqrt {3} \right )\)
\(\mbox {arccot} \left ( \sqrt {3} \right ) = \answer {\frac {\pi }{6}}\)
\(\mbox {arcsec} \left ( 2 \right )\)
\(\mbox {arcsec} \left ( 2 \right ) = \frac {\pi }{3}\)
\(\mbox {arccsc} \left ( 2 \right )\)
\(\mbox {arccsc} \left ( 2 \right ) = \answer {\frac {\pi }{6}}\)
\(\mbox {arcsec} \left ( \sqrt {2} \right )\)
\(\mbox {arcsec} \left ( \sqrt {2} \right ) = \frac {\pi }{4}\)
\(\mbox {arccsc} \left ( \sqrt {2} \right )\)
\(\mbox {arccsc} \left ( \sqrt {2} \right ) = \answer {\frac {\pi }{4}}\)
\(\mbox {arcsec} \left ( \frac {2\sqrt {3}}{3} \right )\)
\(\mbox {arcsec} \left ( \frac {2\sqrt {3}}{3} \right ) = \frac {\pi }{6}\)
\(\mbox {arccsc} \left ( \frac {2\sqrt {3}}{3} \right )\)
\(\mbox {arccsc} \left ( \frac {2\sqrt {3}}{3} \right ) = \answer {\frac {\pi }{3}}\)
\(\mbox {arcsec} \left ( 1 \right )\)
\(\mbox {arcsec} \left ( 1 \right ) = 0\)
\(\mbox {arccsc} \left ( 1 \right )\)
\(\mbox {arccsc} \left ( 1 \right ) = \answer {\frac {\pi }{2}}\)
In Exercises trigfriendexactfirst - trigfriendexactlast , assume that the range of arcsecant is \(\left [0, \frac {\pi }{2} \right ) \cup \left ( \frac {\pi }{2}, \pi \right ]\) and that the range of arccosecant is \(\left [ -\frac {\pi }{2}, 0 \right ) \cup \left (0, \frac {\pi }{2} \right ]\) when finding the exact value.
(See Section arcsecanttrigfriendly .)
\(\mbox {arcsec} \left ( -2 \right )\)
\(\mbox {arcsec} \left ( -2 \right ) = \frac {2\pi }{3}\)
\(\mbox {arcsec} \left ( -\sqrt {2} \right )\)
\(\mbox {arcsec} \left ( -\sqrt {2} \right ) = \answer {\frac {3\pi }{4}}\)
\(\mbox {arcsec} \left ( -\frac {2\sqrt {3}}{3} \right )\)
\(\mbox {arcsec} \left ( -\frac {2\sqrt {3}}{3} \right ) = \frac {5\pi }{6}\)
\(\mbox {arcsec} \left ( -1 \right )\)
\(\mbox {arcsec} \left ( -1 \right ) = \answer {\pi }\)
\(\mbox {arccsc} \left ( -2 \right )\)
\(\mbox {arccsc} \left ( -2 \right ) = -\frac {\pi }{6}\)
\(\mbox {arccsc} \left ( -\sqrt {2} \right )\)
\(\mbox {arccsc} \left ( -\sqrt {2} \right ) = \answer {-\frac {\pi }{4}}\)
\(\mbox {arccsc} \left ( -\frac {2\sqrt {3}}{3} \right )\)
\(\mbox {arccsc} \left ( -\frac {2\sqrt {3}}{3} \right ) = -\frac {\pi }{3}\)
\(\mbox {arccsc} \left ( -1 \right )\)
\(\mbox {arccsc} \left ( -1 \right ) = \answer {-\frac {\pi }{2}}\)
In Exercises
calcfriendexactfirst -
calcfriendexactlast , assume that the range of arcsecant is
\(\left [0, \frac {\pi }{2} \right ) \cup \left [\pi , \frac {3\pi }{2} \right )\) and that the range of arccosecant is
\(\left (0, \frac {\pi }{2} \right ] \cup \left ( \pi , \frac {3\pi }{2} \right ]\) when finding the exact value.
(See Section
arcsecantcalcfriendly .)
\(\mbox {arcsec} \left ( -2 \right )\)
\(\mbox {arcsec} \left ( -2 \right ) = \frac {4\pi }{3}\)
\(\mbox {arcsec} \left ( -\sqrt {2} \right )\)
\(\mbox {arcsec} \left ( -\sqrt {2} \right ) = \answer {\frac {5\pi }{4}}\)
\(\mbox {arcsec} \left ( -\frac {2\sqrt {3}}{3} \right )\)
\(\mbox {arcsec} \left ( -\frac {2\sqrt {3}}{3} \right ) = \frac {7\pi }{6}\)
\(\mbox {arcsec} \left ( -1 \right )\)
\(\mbox {arcsec} \left ( -1 \right ) = \answer {\pi }\)
\(\mbox {arccsc} \left ( -2 \right )\)
\(\mbox {arccsc} \left ( -2 \right ) = \frac {7\pi }{6}\)
\(\mbox {arccsc} \left ( -\sqrt {2} \right )\)
\(\mbox {arccsc} \left ( -\sqrt {2} \right ) = \answer {\frac {5\pi }{4}}\)
\(\mbox {arccsc} \left ( -\frac {2\sqrt {3}}{3} \right )\)
\(\mbox {arccsc} \left ( -\frac {2\sqrt {3}}{3} \right ) = \frac {4\pi }{3}\)
\(\mbox {arccsc} \left ( -1 \right )\)
\(\mbox {arccsc} \left ( -1 \right ) = \answer {\frac {3\pi }{2}}\)
In Exercises
comboexactfirst -
comboexactlast , find the exact value or state that it is undefined.
\(\sin \left (\arcsin \left (\frac {1}{2}\right )\right )\)
\(\sin \left (\arcsin \left (\frac {1}{2}\right )\right ) = \frac {1}{2}\)
\(\sin \left (\arcsin \left (-\frac {\sqrt {2}}{2}\right )\right )\)
\(\sin \left (\arcsin \left (-\frac {\sqrt {2}}{2}\right )\right ) = \answer {-\frac {\sqrt {2}}{2}}\)
\(\sin \left (\arcsin \left (\frac {3}{5}\right )\right )\)
\(\sin \left (\arcsin \left (\frac {3}{5}\right )\right ) = \frac {3}{5}\)
\(\sin \left (\arcsin \left (-0.42\right )\right )\)
\(\sin \left (\arcsin \left (-0.42\right )\right ) = \answer {-0.42}\)
\(\sin \left (\arcsin \left (\frac {5}{4}\right )\right )\)
\(\sin \left (\arcsin \left (\frac {5}{4}\right )\right )\) is undefined.
\(\cos \left (\arccos \left (\frac {\sqrt {2}}{2}\right )\right )\)
\(\cos \left (\arccos \left (\frac {\sqrt {2}}{2}\right )\right ) = \answer {\frac {\sqrt {2}}{2}}\)
\(\cos \left (\arccos \left (-\frac {1}{2}\right )\right )\)
\(\cos \left (\arccos \left (-\frac {1}{2}\right )\right ) = -\frac {1}{2}\)
\(\cos \left (\arccos \left (\frac {5}{13}\right )\right )\)
\(\cos \left (\arccos \left (\frac {5}{13}\right )\right ) = \answer {\frac {5}{13}}\)
\(\cos \left (\arccos \left (-0.998\right )\right )\)
\(\cos \left (\arccos \left (-0.998\right )\right ) = -0.998\)
\(\cos \left (\arccos \left (\pi \right )\right )\)
\(\cos \left (\arccos \left (\pi \right )\right )\) is undefined.
\(\tan \left (\arctan \left (-1\right )\right )\)
\(\tan \left (\arctan \left (-1\right )\right ) = -1\)
\(\tan \left (\arctan \left (\sqrt {3}\right )\right )\)
\(\tan \left (\arctan \left (\sqrt {3}\right )\right ) = \answer {\sqrt {3}}\)
\(\tan \left (\arctan \left (\frac {5}{12}\right )\right )\)
\(\tan \left (\arctan \left (\frac {5}{12}\right )\right ) = \frac {5}{12}\)
\(\tan \left (\arctan \left (0.965\right )\right )\)
\(\tan \left (\arctan \left (0.965\right )\right ) = \answer {0.965}\)
\(\tan \left (\arctan \left ( 3\pi \right )\right )\)
\(\tan \left (\arctan \left ( 3\pi \right )\right ) = 3\pi \)
\(\cot \left (\text {arccot}\left (1\right )\right )\)
\(\cot \left (\text {arccot}\left (1\right )\right ) = \answer {1}\)
\(\cot \left (\text {arccot}\left (-\sqrt {3}\right )\right )\)
\(\cot \left (\text {arccot}\left (-\sqrt {3}\right )\right ) = -\sqrt {3}\)
\(\cot \left (\text {arccot}\left (-\frac {7}{24}\right )\right )\)
\(\cot \left (\text {arccot}\left (-\frac {7}{24}\right )\right ) = \answer {-\frac {7}{24}}\)
\(\cot \left (\text {arccot}\left (-0.001\right )\right )\)
\(\cot \left (\text {arccot}\left (-0.001\right )\right ) = -0.001\)
\(\cot \left (\text {arccot}\left ( \frac {17\pi }{4} \right )\right )\)
\(\cot \left (\text {arccot}\left ( \frac {17\pi }{4} \right )\right ) = \answer {\frac {17\pi }{4}}\)
\(\sec \left (\text {arcsec}\left (2\right )\right )\)
\(\sec \left (\text {arcsec}\left (2\right )\right ) = 2\)
\(\sec \left (\text {arcsec}\left (-1\right )\right )\)
\(\sec \left (\text {arcsec}\left (-1\right )\right ) = \answer {-1}\)
\(\sec \left (\text {arcsec}\left (\frac {1}{2}\right )\right )\)
\(\sec \left (\text {arcsec}\left (\frac {1}{2}\right )\right )\) is undefined.
\(\sec \left (\text {arcsec}\left (0.75\right )\right )\)
\(\sec \left (\text {arcsec}\left (0.75\right )\right )\) is undefined.
\(\sec \left (\text {arcsec}\left ( 117\pi \right )\right )\)
\(\sec \left (\text {arcsec}\left ( 117\pi \right )\right )= 117\pi \)
\(\csc \left (\text {arccsc}\left (\sqrt {2}\right )\right )\)
\(\csc \left (\text {arccsc}\left (\sqrt {2}\right )\right ) = \answer {\sqrt {2}}\)
\(\csc \left (\text {arccsc}\left (-\frac {2\sqrt {3}}{3}\right )\right )\)
\(\csc \left (\text {arccsc}\left (-\frac {2\sqrt {3}}{3}\right )\right ) = -\frac {2\sqrt {3}}{3}\)
\(\csc \left (\text {arccsc}\left (\frac {\sqrt {2}}{2}\right )\right )\)
\(\csc \left (\text {arccsc}\left (\frac {\sqrt {2}}{2}\right )\right )\) is undefined.
\(\csc \left (\text {arccsc}\left (1.0001\right )\right )\)
\(\csc \left (\text {arccsc}\left (1.0001\right )\right ) = 1.0001\)
\(\csc \left (\text {arccsc}\left ( \frac {\pi }{4} \right )\right )\)
\(\csc \left (\text {arccsc}\left ( \frac {\pi }{4} \right )\right )\) is undefined.
In Exercises
morecomboexactfirst -
morecomboexactlast , find the exact value or state that it is undefined.
\(\arcsin \left (\sin \left (\frac {\pi }{6}\right ) \right )\)
\(\arcsin \left (\sin \left (\frac {\pi }{6}\right ) \right ) = \frac {\pi }{6}\)
\(\arcsin \left (\sin \left (-\frac {\pi }{3}\right ) \right )\)
\(\arcsin \left (\sin \left (-\frac {\pi }{3}\right ) \right ) = \answer {-\frac {\pi }{3}}\)
\(\arcsin \left (\sin \left (\frac {3\pi }{4}\right ) \right )\)
\(\arcsin \left (\sin \left (\frac {3\pi }{4}\right ) \right ) = \frac {\pi }{4}\)
\(\arcsin \left (\sin \left (\frac {11\pi }{6}\right ) \right )\)
\(\arcsin \left (\sin \left (\frac {11\pi }{6}\right ) \right ) = \answer {-\frac {\pi }{6}}\)
\(\arcsin \left (\sin \left (\frac {4\pi }{3}\right ) \right )\)
\(\arcsin \left (\sin \left (\frac {4\pi }{3}\right ) \right ) = -\frac {\pi }{3}\)
\(\arccos \left (\cos \left (\frac {\pi }{4}\right ) \right )\)
\(\arccos \left (\cos \left (\frac {\pi }{4}\right ) \right ) = \answer {\frac {\pi }{4}}\)
\(\arccos \left (\cos \left (\frac {2\pi }{3}\right ) \right )\)
\(\arccos \left (\cos \left (\frac {2\pi }{3}\right ) \right ) = \frac {2\pi }{3}\)
\(\arccos \left (\cos \left (\frac {3\pi }{2}\right ) \right )\)
\(\arccos \left (\cos \left (\frac {3\pi }{2}\right ) \right ) = \answer {\frac {\pi }{2}}\)
\(\arccos \left (\cos \left (-\frac {\pi }{6}\right ) \right )\)
\(\arccos \left (\cos \left (-\frac {\pi }{6}\right ) \right ) = \frac {\pi }{6}\)
\(\arccos \left (\cos \left (\frac {5\pi }{4}\right ) \right )\)
\(\arccos \left (\cos \left (\frac {5\pi }{4}\right ) \right ) = \answer {\frac {3\pi }{4}}\)
\(\arctan \left (\tan \left (\frac {\pi }{3}\right ) \right )\)
\(\arctan \left (\tan \left (\frac {\pi }{3}\right ) \right ) = \frac {\pi }{3}\)
\(\arctan \left (\tan \left (-\frac {\pi }{4}\right ) \right )\)
\(\arctan \left (\tan \left (-\frac {\pi }{4}\right ) \right ) = \answer {-\frac {\pi }{4}}\)
\(\arctan \left (\tan \left (\pi \right ) \right )\)
\(\arctan \left (\tan \left (\pi \right ) \right ) = 0\)
\(\arctan \left (\tan \left (\frac {\pi }{2}\right ) \right )\)
\(\arctan \left (\tan \left (\frac {\pi }{2}\right ) \right )\) is undefined
\(\arctan \left (\tan \left (\frac {2\pi }{3}\right ) \right )\)
\(\arctan \left (\tan \left (\frac {2\pi }{3}\right ) \right ) = -\frac {\pi }{3}\)
\(\text {arccot}\left (\cot \left (\frac {\pi }{3}\right ) \right )\)
\(\text {arccot}\left (\cot \left (\frac {\pi }{3}\right ) \right ) = \answer {\frac {\pi }{3}}\)
\(\text {arccot}\left (\cot \left (-\frac {\pi }{4}\right ) \right )\)
\(\text {arccot}\left (\cot \left (-\frac {\pi }{4}\right ) \right ) = \frac {3\pi }{4}\)
\(\text {arccot}\left (\cot \left (\pi \right ) \right )\)
\(\text {arccot}\left (\cot \left (\pi \right ) \right )\) is undefined
\(\text {arccot}\left (\cot \left (\frac {3\pi }{2}\right ) \right )\)
\(\text {arccot}\left (\cot \left (\frac {3\pi }{2}\right ) \right ) = \frac {\pi }{2}\)
\(\text {arccot}\left (\cot \left (\frac {2\pi }{3}\right ) \right )\)
\(\text {arccot}\left (\cot \left (\frac {2\pi }{3}\right ) \right ) = \answer {\frac {2\pi }{3}}\)
In Exercises moreextracombofirst - moreextracombolast , assume that the range of arcsecant is \(\left [0, \frac {\pi }{2} \right ) \cup \left ( \frac {\pi }{2}, \pi \right ]\) and that the range of arccosecant is \(\left [ -\frac {\pi }{2}, 0 \right ) \cup \left (0, \frac {\pi }{2} \right ]\) when finding the exact value.
(See Section arcsecanttrigfriendly .)
\(\text {arcsec}\left (\sec \left (\frac {\pi }{4}\right ) \right )\)
\(\text {arcsec}\left (\sec \left (\frac {\pi }{4}\right ) \right ) = \frac {\pi }{4}\)
\(\text {arcsec}\left (\sec \left (\frac {4\pi }{3}\right ) \right )\)
\(\text {arcsec}\left (\sec \left (\frac {4\pi }{3}\right ) \right ) = \answer {\frac {2\pi }{3}}\)
\(\text {arcsec}\left (\sec \left ( \frac {5\pi }{6} \right ) \right )\)
\(\text {arcsec}\left (\sec \left ( \frac {5\pi }{6} \right ) \right ) = \frac {5\pi }{6}\)
\(\text {arcsec}\left (\sec \left (-\frac {\pi }{2} \right ) \right )\)
\(\text {arcsec}\left (\sec \left (-\frac {\pi }{2} \right ) \right )\) is undefined.
\(\text {arcsec}\left (\sec \left (\frac {5\pi }{3}\right ) \right )\)
\(\text {arcsec}\left (\sec \left (\frac {5\pi }{3}\right ) \right ) = \frac {\pi }{3}\)
\(\text {arccsc}\left (\csc \left (\frac {\pi }{6}\right ) \right )\)
\(\text {arccsc}\left (\csc \left (\frac {\pi }{6}\right ) \right ) = \answer {\frac {\pi }{6}}\)
\(\text {arccsc}\left (\csc \left (\frac {5\pi }{4}\right ) \right )\)
\(\text {arccsc}\left (\csc \left (\frac {5\pi }{4}\right ) \right ) = -\frac {\pi }{4}\)
\(\text {arccsc}\left (\csc \left ( \frac {2\pi }{3} \right ) \right )\)
\(\text {arccsc}\left (\csc \left ( \frac {2\pi }{3} \right ) \right ) = \answer {\frac {\pi }{3}}\)
\(\text {arccsc}\left (\csc \left (-\frac {\pi }{2} \right ) \right )\)
\(\text {arccsc}\left (\csc \left (-\frac {\pi }{2} \right ) \right ) = -\frac {\pi }{2}\)
\(\text {arccsc}\left (\csc \left (\frac {11\pi }{6}\right ) \right )\)
\(\text {arccsc}\left (\csc \left (\frac {11\pi }{6}\right ) \right ) = \answer {-\frac {\pi }{6}}\)
\(\text {arcsec}\left (\sec \left (\frac {11\pi }{12}\right ) \right )\)
\(\text {arcsec}\left (\sec \left (\frac {11\pi }{12}\right ) \right ) = \frac {11\pi }{12}\)
\(\text {arccsc}\left (\csc \left (\frac {9\pi }{8}\right ) \right )\)
\(\text {arccsc}\left (\csc \left (\frac {9\pi }{8}\right ) \right ) = \answer {-\frac {\pi }{8}}\)
In Exercises
extracombofirst -
extracombolast , assume that the range of arcsecant is
\(\left [0, \frac {\pi }{2} \right ) \cup \left [\pi , \frac {3\pi }{2} \right )\) and that the range of arccosecant is
\(\left (0, \frac {\pi }{2} \right ] \cup \left ( \pi , \frac {3\pi }{2} \right ]\) when finding the exact value.
(See Section
arcsecantcalcfriendly .)
\(\text {arcsec}\left (\sec \left (\frac {\pi }{4}\right ) \right )\)
\(\text {arcsec}\left (\sec \left (\frac {\pi }{4}\right ) \right ) = \frac {\pi }{4}\)
\(\text {arcsec}\left (\sec \left (\frac {4\pi }{3}\right ) \right )\)
\(\text {arcsec}\left (\sec \left (\frac {4\pi }{3}\right ) \right ) = \answer {\frac {4\pi }{3}}\)
\(\text {arcsec}\left (\sec \left ( \frac {5\pi }{6} \right ) \right )\)
\(\text {arcsec}\left (\sec \left ( \frac {5\pi }{6} \right ) \right ) = \frac {7\pi }{6}\)
\(\text {arcsec}\left (\sec \left (-\frac {\pi }{2} \right ) \right )\)
\(\text {arcsec}\left (\sec \left (-\frac {\pi }{2} \right ) \right )\) is undefined.
\(\text {arcsec}\left (\sec \left (\frac {5\pi }{3}\right ) \right )\)
\(\text {arcsec}\left (\sec \left (\frac {5\pi }{3}\right ) \right ) = \frac {\pi }{3}\)
\(\text {arccsc}\left (\csc \left (\frac {\pi }{6}\right ) \right )\)
\(\text {arccsc}\left (\csc \left (\frac {\pi }{6}\right ) \right ) = \answer {\frac {\pi }{6}}\)
\(\text {arccsc}\left (\csc \left (\frac {5\pi }{4}\right ) \right )\)
\(\text {arccsc}\left (\csc \left (\frac {5\pi }{4}\right ) \right ) = \frac {5\pi }{4}\)
\(\text {arccsc}\left (\csc \left ( \frac {2\pi }{3} \right ) \right )\)
\(\text {arccsc}\left (\csc \left ( \frac {2\pi }{3} \right ) \right ) = \answer {\frac {\pi }{3}}\)
\(\text {arccsc}\left (\csc \left (-\frac {\pi }{2} \right ) \right )\)
\(\text {arccsc}\left (\csc \left (-\frac {\pi }{2} \right ) \right ) = \frac {3\pi }{2}\)
\(\text {arccsc}\left (\csc \left (\frac {11\pi }{6}\right ) \right )\)
\(\text {arccsc}\left (\csc \left (\frac {11\pi }{6}\right ) \right ) = \answer {\frac {7\pi }{6}}\)
\(\text {arcsec}\left (\sec \left (\frac {11\pi }{12}\right ) \right )\)
\(\text {arcsec}\left (\sec \left (\frac {11\pi }{12}\right ) \right ) = \frac {13\pi }{12}\)
\(\text {arccsc}\left (\csc \left (\frac {9\pi }{8}\right ) \right )\)
\(\text {arccsc}\left (\csc \left (\frac {9\pi }{8}\right ) \right ) = \answer {\frac {9\pi }{8}}\)
In Exercises
stillmoreexactfirst -
stillmoreexactlast , find the exact value or state that it is undefined.
\(\sin \left (\arccos \left (-\frac {1}{2}\right )\right )\)
\(\sin \left (\arccos \left (-\frac {1}{2}\right )\right ) = \frac {\sqrt {3}}{2}\)
\(\sin \left (\arccos \left (\frac {3}{5}\right )\right )\)
\(\sin \left (\arccos \left (\frac {3}{5}\right )\right ) = \answer {\frac {4}{5}}\)
\(\sin \left (\arctan \left (-2\right )\right )\)
\(\sin \left (\arctan \left (-2\right )\right ) = -\frac {2\sqrt {5}}{5}\)
\(\sin \left (\text {arccot}\left (\sqrt {5}\right )\right )\)
\(\sin \left (\text {arccot}\left (\sqrt {5}\right )\right ) = \answer {\frac {\sqrt {6}}{6}}\)
\(\sin \left (\text {arccsc}\left (-3\right )\right )\)
\(\sin \left (\text {arccsc}\left (-3\right )\right ) = -\frac {1}{3}\)
\(\cos \left (\arcsin \left (-\frac {5}{13}\right )\right )\)
\(\cos \left (\arcsin \left (-\frac {5}{13}\right )\right ) = \answer {\frac {12}{13}}\)
\(\cos \left (\arctan \left (\sqrt {7} \right )\right )\)
\(\cos \left (\arctan \left (\sqrt {7} \right )\right ) = \frac {\sqrt {2}}{4}\)
\(\cos \left (\text {arccot}\left ( 3 \right )\right )\)
\(\cos \left (\text {arccot}\left ( 3 \right )\right ) = \answer {\frac {3\sqrt {10}}{10}}\)
\(\cos \left (\text {arcsec}\left ( 5 \right )\right )\)
\(\cos \left (\text {arcsec}\left ( 5 \right )\right ) = \frac {1}{5}\)
\(\tan \left (\arcsin \left (-\frac {2\sqrt {5}}{5}\right )\right )\)
\(\tan \left (\arcsin \left (-\frac {2\sqrt {5}}{5}\right )\right )=\answer {-2}\)
\(\tan \left (\arccos \left (-\frac {1}{2}\right )\right )\)
\(\tan \left (\arccos \left (-\frac {1}{2}\right )\right ) = -\sqrt {3}\)
\(\tan \left (\text {arcsec}\left (\frac {5}{3}\right )\right )\)
\(\tan \left (\text {arcsec}\left (\frac {5}{3}\right )\right ) = \answer {\frac {4}{3}}\)
\(\tan \left (\text {arccot}\left ( 12 \right )\right )\)
\(\tan \left (\text {arccot}\left ( 12 \right )\right ) = \frac {1}{12}\)
\(\cot \left (\arcsin \left (\frac {12}{13}\right )\right )\)
\(\cot \left (\arcsin \left (\frac {12}{13}\right )\right ) = \answer {\frac {5}{12}}\)
\(\cot \left (\arccos \left (\frac {\sqrt {3}}{2}\right )\right )\)
\(\cot \left (\arccos \left (\frac {\sqrt {3}}{2}\right )\right ) = \sqrt {3}\)
\(\cot \left (\text {arccsc}\left (\sqrt {5}\right )\right )\)
\(\cot \left (\text {arccsc}\left (\sqrt {5}\right )\right ) = \answer {2}\)
\(\cot \left (\arctan \left ( 0.25 \right )\right )\)
\(\cot \left (\arctan \left ( 0.25 \right )\right ) = 4\)
\(\sec \left (\arccos \left (\frac {\sqrt {3}}{2}\right )\right )\)
\(\sec \left (\arccos \left (\frac {\sqrt {3}}{2}\right )\right ) = \answer {\frac {2\sqrt {3}}{3}}\)
\(\sec \left (\arcsin \left (-\frac {12}{13}\right )\right )\)
\(\sec \left (\arcsin \left (-\frac {12}{13}\right )\right ) = \frac {13}{5}\)
\(\sec \left (\arctan \left (10\right )\right )\)
\(\sec \left (\arctan \left (10\right )\right ) = \answer {\sqrt {101}}\)
\(\sec \left (\text {arccot}\left (-\frac {\sqrt {10}}{10}\right )\right )\)
\(\sec \left (\text {arccot}\left (-\frac {\sqrt {10}}{10}\right )\right ) = -\sqrt {11}\)
\(\csc \left (\text {arccot}\left (9 \right )\right )\)
\(\csc \left (\text {arccot}\left (9 \right )\right ) = \answer {\sqrt {82}}\)
\(\csc \left (\arcsin \left (\frac {3}{5}\right )\right )\)
\(\csc \left (\arcsin \left (\frac {3}{5}\right )\right ) = \frac {5}{3}\)
\(\csc \left (\arctan \left (-\frac {2}{3}\right )\right )\)
\(\csc \left (\arctan \left (-\frac {2}{3}\right )\right ) = \answer {-\frac {\sqrt {13}}{2}}\)
In Exercises
exactvalueidenfirst -
exactvalueidenlast , find the exact value or state that it is undefined.
\(\sin \left (\arcsin \left ( \frac {5}{13} \right ) + \frac {\pi }{4}\right )\)
\(\sin \left (\arcsin \left ( \frac {5}{13} \right ) + \frac {\pi }{4}\right ) = \frac {17\sqrt {2}}{26}\)
\(\cos \left ( \text {arcsec}(3) + \arctan (2) \right )\)
\(\cos \left ( \text {arcsec}(3) + \arctan (2) \right ) = \answer {\frac {\sqrt {5} - 4\sqrt {10}}{15}}\)
\(\tan \left ( \arctan (3) + \arccos \left (-\frac {3}{5}\right ) \right )\)
\(\tan \left ( \arctan (3) + \arccos \left (-\frac {3}{5}\right ) \right ) = \frac {1}{3}\)
\(\sin \left (2 \arcsin \left (-\frac {4}{5}\right )\right )\)
\(\sin \left (2 \arcsin \left (-\frac {4}{5}\right )\right )= \answer {-\frac {24}{25}}\)
\(\sin \left (2\text {arccsc}\left (\frac {13}{5}\right )\right )\)
\(\sin \left (2\text {arccsc}\left (\frac {13}{5}\right )\right ) = \frac {120}{169}\)
\(\sin \left (2 \arctan \left (2\right )\right )\)
\(\sin \left (2 \arctan \left (2\right )\right ) = \frac {4}{5}\)
\(\cos \left (2 \arcsin \left (\frac {3}{5}\right )\right )\)
\(\cos \left (2 \arcsin \left (\frac {3}{5}\right )\right ) = \frac {7}{25}\)
\(\cos \left (2 \text {arcsec}\left (\frac {25}{7}\right )\right )\)
\(\cos \left (2 \text {arcsec}\left (\frac {25}{7}\right )\right ) = \answer {-\frac {527}{625}}\)
\(\cos \left (2 \text {arccot}\left (-\sqrt {5}\right )\right )\)
\(\cos \left (2 \text {arccot}\left (-\sqrt {5}\right )\right ) = \frac {2}{3}\)
\(\sin \left ( \frac {\arctan (2)}{2} \right )\)
\(\sin \left ( \frac {\arctan (2)}{2} \right ) = \answer {\sqrt {\frac {5-\sqrt {5}}{10}}}\)
In Exercises rewritefirst - rewritelast , rewrite each of the following composite functions as algebraic functions of \(x\) and state the domain.
\(f(x) = \sin \left ( \arccos \left ( x \right ) \right )\)
\(f(x) = \sin \left ( \arccos \left ( x \right ) \right ) = \sqrt {1 - x^{2}}\) for \(-1 \leq x \leq 1\)
\(f(x) = \cos \left ( \arctan \left ( x \right ) \right )\)
\(f(x) = \cos \left ( \arctan \left ( x \right ) \right ) = \frac {1}{\sqrt {1 + x^{2}}}\) for all \(x\)
\(f(x) = \tan \left ( \arcsin \left ( x \right ) \right )\)
\(f(x) =\tan \left ( \arcsin \left ( x \right ) \right ) = \frac {x}{\sqrt {1 - x^{2}}}\) for \(-1 < x < 1\)
\(f(x) = \sec \left ( \arctan \left ( x \right ) \right )\)
\(f(x) =\sec \left ( \arctan \left ( x \right ) \right ) = \sqrt {1 + x^{2}}\) for all \(x\)
\(f(x) = \csc \left ( \arccos \left ( x \right ) \right )\)
\(f(x) =\csc \left ( \arccos \left ( x \right ) \right ) = \frac {1}{\sqrt {1 - x^{2}}}\) for \(-1 < x < 1\)
\(f(x) = \sin \left ( 2 \arctan \left ( x \right ) \right )\)
\(f(x) =\sin \left ( 2 \arctan \left ( x \right ) \right ) = \frac {2x}{x^{2} + 1}\) for all \(x\)
\(f(x) = \sin \left ( 2 \arccos \left ( x \right ) \right )\)
\(f(x) =\sin \left ( 2 \arccos \left ( x \right ) \right ) = 2x\sqrt {1-x^2}\) for \(-1 \leq x \leq 1\)
\(f(x) = \cos \left ( 2 \arctan \left ( x \right ) \right )\)
\(f(x) =\cos \left ( 2 \arctan \left ( x \right ) \right ) = \frac {1 - x^{2}}{1 + x^{2}}\) for all \(x\)
\(f(x) = \sin (\arccos (2x))\)
\(f(x) =\sin (\arccos (2x)) = \sqrt {1-4x^2}\) for \(-\frac {1}{2} \leq x \leq \frac {1}{2}\)
\(f(x) = \sin \left (\arccos \left (\frac {x}{5}\right )\right )\)
\(f(x) =\sin \left (\arccos \left (\frac {x}{5}\right )\right ) = \frac {\sqrt {25-x^2}}{5}\) for \(-5 \leq x \leq 5\)
\(f(x) = \cos \left (\arcsin \left (\frac {x}{2}\right )\right )\)
\(f(x) =\cos \left (\arcsin \left (\frac {x}{2}\right )\right ) = \frac {\sqrt {4-x^2}}{2}\) for \(-2 \leq x \leq 2\)
\(f(x) = \cos \left (\arctan \left (3x\right )\right )\)
\(f(x) =\cos \left (\arctan \left (3x\right )\right ) = \frac {1}{\sqrt {1+9x^{2}}}\) for all \(x\)
\(f(x) = \sin (2 \arcsin (7x))\)
\(f(x) =\sin (2 \arcsin (7x)) = 14x \sqrt {1-49x^2}\) for \(-\frac {1}{7} \leq x \leq \frac {1}{7}\)
\(f(x) = \sin \left (2 \arcsin \left ( \frac {x\sqrt {3}}{3} \right ) \right )\)
\(f(x) =\sin \left (2 \arcsin \left ( \frac {x\sqrt {3}}{3} \right ) \right ) = \frac {2x\sqrt {3-x^2}}{3}\) for \(-\sqrt {3} \leq x \leq \sqrt {3}\)
\(f(x) = \cos (2 \arcsin (4x))\)
\(f(x) =\cos (2 \arcsin (4x)) = 1 - 32x^2\) for \(-\frac {1}{4} \leq x \leq \frac {1}{4}\)
\(f(x) = \sec (\arctan (2x))\tan (\arctan (2x))\)
\(f(x) =\sec (\arctan (2x))\tan (\arctan (2x)) = 2x \sqrt {1+4x^2}\) for all \(x\)
\(f(x) = \sin \left ( \arcsin (x) + \arccos (x) \right )\)
\(f(x) =\sin \left ( \arcsin (x) + \arccos (x) \right ) = 1\) for \(-1 \leq x \leq 1\)
\(f(x) = \cos \left ( \arcsin (x) + \arctan (x) \right )\)
\(f(x) =\cos \left ( \arcsin (x) + \arctan (x) \right ) = \frac {\sqrt {1 - x^{2}} - x^{2}}{\sqrt {1 + x^{2}}}\) for \(-1 \leq x \leq 1\)
\(f(x) = \tan \left ( 2 \arcsin (x) \right )\)
\(f(x) =\tan \left ( 2 \arcsin (x) \right ) = \frac {2x\sqrt {1 - x^{2}}}{1 - 2x^{2}}\) for [1] \(x\) in \(\left (-1, -\frac {\sqrt {2}}{2}\right ) \cup \left (-\frac {\sqrt {2}}{2}, \frac {\sqrt {2}}{2} \right ) \cup \left (\frac {\sqrt {2}}{2}, 1\right )\)
\(f(x) = \sin \left ( \frac {1}{2}\arctan (x) \right )\)
\(f(x) =\sin \left ( \frac {1}{2}\arctan (x) \right ) = \left \{ \begin{array}{rr} \sqrt {\frac {\sqrt {x^{2} + 1} - 1}{2\sqrt {x^{2} + 1}}} & \text {for $x \geq 0$} \\ [10pt] -\sqrt {\frac {\sqrt {x^{2} + 1} - 1}{2\sqrt {x^{2} + 1}}} & \text {for $x < 0$} \end{array}\right . \)
If
\(\theta = \arcsin \left (\frac {x}{2}\right )\) , find an expression for
\(\theta + \sin (2\theta )\) in terms of
\(x\) .
\(\theta + \sin (2\theta ) = \arcsin \left ( \frac {x}{2} \right ) + \frac {x\sqrt {4 - x^{2}}}{2}\)
If
\(\theta = \arctan \left (\frac {x}{7}\right )\) , find an expression for
\(\frac {1}{2}\theta - \frac {1}{2}\sin (2\theta )\) in terms of
\(x\) .
\(\frac {1}{2}\theta - \frac {1}{2}\sin (2\theta ) = \frac {1}{2} \arctan \left ( \frac {x}{7} \right ) - \frac {7x}{x^{2} + 49}\)
If
\(\theta = \mbox {arcsec}\left (\frac {x}{4}\right )\) , find an expression for
\(4\tan (\theta ) - 4\theta \) in terms of
\(x\) assuming
\(x \geq 4\) .
\(4\tan (\theta ) - 4\theta = \sqrt {x^{2} - 16} - 4\mbox {arcsec} \left ( \frac {x}{4} \right )\)
In Exercises
equarctrigfirst -
equarctriglast , solve the equation using the techniques discussed in Example
basicinverseeqns then approximate the solutions which lie in
the interval
\([0, 2\pi )\) to four decimal places.
\(\sin (\theta ) = \frac {7}{11}\)
\(\theta = \arcsin \left (\frac {7}{11}\right ) + 2\pi k\) or \(\theta = \pi - \arcsin \left (\frac {7}{11}\right ) + 2\pi k\) , in \([0, 2\pi )\) , \(\theta \approx 0.6898, \, 2.4518\)
\(\cos (\theta ) = -\frac {2}{9}\)
\(\theta = \arccos \left (-\frac {2}{9}\right ) + 2\pi k\) or \(\theta = - \arccos \left (-\frac {2}{9}\right ) + 2\pi k\) , in \([0, 2\pi )\) , \(\theta \approx 1.7949, \, 4.4883\)
\(\sin (\theta ) = -0.569\)
\(\theta = \pi + \arcsin (0.569) + 2\pi k\) or \(\theta = 2\pi - \arcsin (0.569) + 2\pi k\) , in \([0, 2\pi )\) , \(\theta \approx 3.7469, \, 5.6779\)
\(\cos (\theta ) = 0.117\)
\(\theta = \arccos (0.117) + 2\pi k\) or \(\theta = 2\pi - \arccos (0.117) + 2\pi k\) , in \([0, 2\pi )\) , \(\theta \approx 1.4535, \, 4.8297\)
\(\sin (\theta ) = 0.008\)
\(\theta = \arcsin (0.008) + 2\pi k\) or \(\theta = \pi - \arcsin (0.008) + 2\pi k\) , in \([0, 2\pi )\) , \(\theta \approx 0.0080, \, 3.1336\)
\(\cos (\theta ) = \frac {359}{360}\)
\(\theta = \arccos \left (\frac {359}{360}\right ) + 2\pi k\) or \(\theta = 2\pi - \arccos \left (\frac {359}{360}\right ) + 2\pi k\) , in \([0, 2\pi )\) , \(\theta \approx 0.0746, \, 6.2086\)
\(\tan (t) = 117\)
\(t = \arctan (117) + \pi k\) , in \([0, 2\pi )\) , \(t \approx 1.56225, \, 4.70384\) \(t = \arctan \left (-\frac {1}{12}\right ) + \pi k\) , in \([0, 2\pi )\) , \(t \approx 3.0585, \, 6.2000\)
\(\cot (t) = -12\)
\(t = \arctan \left (-\frac {1}{12}\right ) + \pi k\) , in \([0, 2\pi )\) , \(t \approx 3.0585, \, 6.2000\)
\(\sec (t) = \frac {3}{2}\)
\(t = \arccos \left (\frac {2}{3}\right ) + 2\pi k\) or \(t = 2\pi - \arccos \left (\frac {2}{3}\right ) + 2\pi k\) , in \([0, 2\pi )\) , \(t \approx 0.8411, \, 5.4422\)
\(\csc (t) = -\frac {90}{17}\)
\(t = \pi + \arcsin \left (\frac {17}{90}\right ) + 2\pi k\) or \(t = 2\pi - \arcsin \left (\frac {17}{90}\right ) + 2\pi k\) , in \([0, 2\pi )\) , \(t \approx 3.3316, \, 6.0932\)
\(\tan (t) = -\sqrt {10}\)
\(t = \arctan \left (-\sqrt {10}\right ) + \pi k\) , in \([0, 2\pi )\) , \(t \approx 1.8771, \, 5.0187\)
\(\sin (t) = \frac {3}{8}\)
\(t = \arcsin \left (\frac {3}{8}\right ) + 2\pi k\) or \(t = \pi - \arcsin \left (\frac {3}{8}\right ) + 2\pi k\) , in \([0, 2\pi )\) , \(t \approx 0.3844, \, 2.7572\)
\(\cos (x) = -\frac {7}{16}\)
\(x = \arccos \left (-\frac {7}{16}\right ) + 2\pi k\) or \(x = - \arccos \left (-\frac {7}{16}\right ) + 2\pi k\) , in \([0, 2\pi )\) , \(x \approx 2.0236, \, 4.2596\)
\(\tan (x) = 0.03\)
\(x = \arctan (0.03) + \pi k\) , in \([0, 2\pi )\) , \(x \approx 0.0300, \, 3.1716\)
\(\sin (x) = 0.3502\)
\(x = \arcsin (0.3502) + 2\pi k\) or \(x = \pi - \arcsin (0.3502) + 2\pi k\) , in \([0, 2\pi )\) , \(x \approx 0.3578, \,2.784\)
\(\sin (x) = -0.721\)
\(x = \pi + \arcsin (0.721) + 2\pi k\) or \(x = 2\pi - \arcsin (0.721) + 2\pi k\) , in \([0, 2\pi )\) , \(x \approx 3.9468, \, 5.4780\)
\(\cos (x) = 0.9824\)
\(x = \arccos (0.9824) + 2\pi k\) or \(x = 2\pi - \arccos (0.9824) + 2\pi k\) , in \([0, 2\pi )\) , \(x \approx 0.1879, \, 6.0953\)
\(\cos (x) = -0.5637\)
\(x = \arccos (-0.5637) + 2\pi k\) or \(x = - \arccos (-0.5637) + 2\pi k\) , in \([0, 2\pi )\) , \(x \approx 2.1697, \, 4.1135\)
\(\cot (x) = \frac {1}{117}\)
\(x = \arctan (117) + \pi k\) , in \([0, 2\pi )\) , \(x \approx 1.5622, \, 4.7038\)
\(\tan (x) = -0.6109\)
\(x = \arctan (-0.6109) + \pi k\) , in \([0, 2\pi )\) , \(x \approx 2.5932, \, 5.7348\)
In Exercises
rewritesinusoidfirst -
rewritesinusoidlast , rewrite the given function as a sinusoid of the form
\(C(t) = A \cos (\omega t + \phi )\) and
\(S(t) = A\sin (\omega t + \phi )\) (See Example
expandedsinusoidinverseex .) Approximate the value of
\(\phi \) (which
is in radians, of course) to four decimal places.
\(f(t) = 5\sin (3t) + 12\cos (3t)\)
\(f(t) = 5\sin (3t) + 12\cos (3t) = 13\sin \left (3t + \arcsin \left (\frac {12}{13}\right )\right ) \approx 13\sin (3t + 1.1760)\)
\(f(t) = 5\sin (3t) + 12\cos (3t) = 13\cos \left (3t + \arcsin \left (-\frac {5}{13}\right )\right ) \approx 13\cos (3t -0.3948)\)
\(f(t) = 3\cos (2t) + 4\sin (2t)\)
\(f(t) = 3\cos (2t) + 4\sin (2t) = 5\sin \left (2t+\arcsin \left (\frac {3}{5}\right ) \right ) \approx 5\sin (2t+0.6435)\)
\(f(t) = 3\cos (2t) + 4\sin (2t) = 5\cos \left (2t+\arcsin \left (-\frac {4}{5}\right ) \right ) \approx 5\cos (2t-0.9273)\)
\(f(t) = \cos (t) - 3\sin (t)\)
\(f(t) = \cos (t) - 3\sin (t) = \sqrt {10} \sin \left (t + \arccos \left (-\frac {3\sqrt {10}}{10} \right )\right ) \approx \sqrt {10} \sin (t + 2.8198)\)
\(f(t) = \cos (t) - 3\sin (t) = \sqrt {10} \cos \left (t + \arcsin \left (\frac {3\sqrt {10}}{10} \right )\right ) \approx \sqrt {10} \cos (t + 1.2490)\)
\(f(t) = 7\sin (10t) - 24\cos (10t)\)
\(f(t) = 7\sin (10t) - 24\cos (10t) = 25\sin \left ( 10t + \arcsin \left (-\frac {24}{25}\right )\right ) \approx 25 \sin (10t-1.2870)\)
\(f(t) = 7\sin (10t) - 24\cos (10t) = 25\cos \left ( 10t + \pi + \arcsin \left (\frac {7}{25}\right )\right ) \approx 25 \cos (10t+3.4254)\)
\(f(t) = -\cos (t) - 2\sqrt {2} \sin (t)\)
\(f(t) = -\cos (t) - 2\sqrt {2} \sin (t) = 3\sin \left (t+\pi + \arcsin \left (\frac {1}{3}\right )\right ) \approx 3\sin (t+3.4814)\)
\(f(t) = -\cos (t) - 2\sqrt {2} \sin (t) = 3\cos \left (t+ \arccos \left (-\frac {1}{3}\right )\right ) \approx 3\sin (t+1.9106)\)
\(f(t) = 2\sin (t) - \cos (t)\)
\(f(t) = 2\sin (t) - \cos (t) = \sqrt {5}\sin \left (t + \arcsin \left (-\frac {\sqrt {5}}{5}\right )\right ) \approx \sqrt {5}\sin (t -0.4636)\)
\(f(t) = 2\sin (t) - \cos (t) = \sqrt {5}\cos \left (t + \pi + \arcsin \left (\frac {2\sqrt {5}}{5}\right )\right ) \approx \sqrt {5}\cos (t + 4.2487)\)
In Exercises domainexerfirst - domainexerlast , find the domain of the given function. Write your answers in interval notation.
\(f(x) = \arcsin (5x)\)
\(\left [-\frac {1}{5}, \frac {1}{5}\right ]\)
\(f(x) = \arccos \left (\frac {3x-1}{2} \right )\)
\(\left [-\frac {1}{3}, 1 \right ]\)
\(f(x) = \arcsin \left (2x^2\right )\)
\(\left [-\frac {\sqrt {2}}{2}, \frac {\sqrt {2}}{2}\right ]\)
\(f(x) = \arccos \left (\frac {1}{x^2-4}\right )\)
\((-\infty , -\sqrt {5}] \cup [-\sqrt {3}, \sqrt {3}] \cup [\sqrt {5}, \infty )\)
\(f(x) = \text {arccot}\left (\frac {2x}{x^2-9}\right )\)
\((-\infty , -3) \cup (-3,3) \cup (3, \infty )\)
\(f(x) =\arctan (\ln (2x-1))\)
\(\left (\frac {1}{2}, \infty \right )\)
\(f(x) = \text {arccot}(\sqrt {2x-1})\)
\(\left [\frac {1}{2}, \infty \right )\)
\(f(x) = \text {arcsec}(12x)\)
\(\left (-\infty , -\frac {1}{12}\right ] \cup \left [\frac {1}{12}, \infty \right )\)
\(f(x) = \text {arccsc}(x+5)\)
\((-\infty , -6] \cup [-4, \infty )\)
\(f(x) = \text {arcsec}\left (\frac {x^3}{8}\right )\)
\((-\infty , -2] \cup [2, \infty )\)
\(f(x) = \text {arccsc}\left (e^{2x}\right )\)
Find the following limits.
\(\ds { \lim _{x \rightarrow 1^{-}} \arcsin (x)}\)
\(\ds { \lim _{x \rightarrow 1^{-}} \arcsin (x)}\) \(=\frac {\pi }{2}\)
\(\ds {\lim _{x \rightarrow -\infty } \arctan (3x)}\)
\(\ds {\lim _{x \rightarrow -\infty } \arctan (3x) = -\infty }\)
\(\ds {\lim _{x \rightarrow 1} \text {arcsec}(2x)}\)
\(\ds {\lim _{x \rightarrow 1} \text {arcsec}(2x)}\) \(= \frac {\pi }{3}\)
Find a nonzero number \(x\) where \(\mbox {arccot}(x) \neq \arctan \left ( \frac {1}{x} \right )\) .
Find an example where \(\mbox {arcsec}(x) \neq \arccos \left ( \frac {1}{x} \right )\) if we use \(\left [0, \frac {\pi }{2} \right ) \cup \left [ \pi , \frac {3\pi }{2} \right )\) as the range of \(f(x) = \mbox {arcsec}(x)\) .
Show that \(\arcsin (x) + \arccos (x) = \frac {\pi }{2}\) for \(-1 \leq x \leq 1\) .
Discuss with your classmates why \(\arcsin \left (\frac {1}{2}\right ) \neq 30\) .
Use the diagram below along with the accompanying questions to show:
\[\arctan (1) + \arctan (2) + \arctan (3) = \pi \]
Clearly \(\triangle AOB\) and \(\triangle BCD\) are right triangles because the line through \(O\) and \(A\) and the line through \(C\) and \(D\) are perpendicular to
the \(x\) -axis. Use the distance formula to show that \(\triangle BAD\) is also a right triangle (with \(\angle BAD\) being the right angle) by showing
that the sides of the triangle satisfy the Pythagorean Theorem.
Use \(\triangle AOB\) to show that \(\alpha = \arctan (1)\)
Use \(\triangle BAD\) to show that \(\beta = \arctan (2)\)
Use \(\triangle BCD\) to show that \(\gamma = \arctan (3)\)
Use the fact that \(O\) , \(B\) and \(C\) all lie on the \(x\) -axis to conclude that \(\alpha + \beta + \gamma = \pi \) . Thus \(\arctan (1) + \arctan (2) + \arctan (3) = \pi \) .
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