The inverse sine, cosine and tangent functions

Notation of Inverse Trig Functions

$\sin {\theta }=r\to \arcsin {r}=\sin ^{-1}{r}=\theta$. The domain of $y=\arcsin {x}$ is $[-1,1]$, and its range is $[{\frac {-\pi }{2}},{\frac {\pi }{2}}]$.

$\cos {\theta }=r\to \arccos {r}=\cos ^{-1}{r}=\theta$. Domain: $[-1,1]$; range: $[0,\pi ]$.

$\tan {\theta }=r\to \arctan {r}=\tan ^{-1}{r}=\theta$. Domain: $(-\infty ,\infty )$; range: $({\frac {-\pi }{2}},{\frac {\pi }{2}})$.

$\csc {\theta }=r\to \operatorname {arccsc} {r}=\csc ^{-1}{r}=\theta$. Domain: $(-\infty ,-1],[1,\infty ]$; range: $[{\frac {-\pi }{2}},{\frac {\pi }{2}}]$.

$\sec {\theta }=r\to \operatorname {arcsec} {r}=\sec ^{-1}{r}=\theta$. Domain: $(-\infty ,-1],[1,\infty ]$; range: $[0,\pi ]$.

$\cot {\theta }=r\to \operatorname {arccot} {r}=\cot ^{-1}{r}=\theta$. Domain: $(-\infty ,\infty )$; range: $(0,\pi )$.

Example: $\sin {\frac {\pi }{4}}={\frac {\sqrt {2}}{2}}$, so $\arcsin {\frac {\sqrt {2}}{2}}={\frac {\pi }{4}}$. Even though $\sin {\frac {3\pi }{4}}={\frac {\sqrt {2}}{2}}$ as well, ${\frac {3\pi }{4}}$ is outside of the range for $\arcsin {x}$.

Notes: $\csc {x}=1/\sin {x}$, and $\operatorname {arccsc} {x}=\arcsin {(1/x)}$. $\operatorname {arccsc} {x}$ does NOT equal $1/\arcsin {x}$. Similarly, $\operatorname {arcsec} {x}=\arccos {(1/x)}$, and NOT $1/\arccos {x}$.

Inverse Trig Functions

$\int \arcsin u\mathrm {d} u=u\arcsin u+{\sqrt {1-u^{2}}}+C$

$\int \arccos u\mathrm {d} u=u\arccos u-{\sqrt {1-u^{2}}}+C$

$\int \arctan u\mathrm {d} u=u\arctan u-\ln {\sqrt {1+u^{2}}}+C$

$\int \operatorname {arccot} u\mathrm {d} u=u\operatorname {arccot} u+\ln {\sqrt {1+u^{2}}}+C$

$\int \operatorname {arcsec} u\mathrm {d} u=u\operatorname {arcsec} u+\ln \left\vert u+{\sqrt {u^{2}-1}}\right\vert +C$

$\int \operatorname {arccsc} u\mathrm {d} u=u\operatorname {arccsc} u+\ln \left\vert u+{\sqrt {u^{2}-1}}\right\vert +C$

Resources

• Intro to Inverse Trig Functions, Khan Academy
• Inverse Trig Functions, Paul's Online Notes