The inverse sine, cosine and tangent functions

Notation of Inverse Trig Functions

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sin{\theta} = r \to \arcsin{r} = \sin^{-1}{r} = \theta} . The domain of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \arcsec{x} = \arccos{(1/x)}} , and NOT Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 1/\arccos{x} } .

Inverse Trig Functions

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int \arcsin u \mathrm{d}u = u \arcsin u + \sqrt {1 - u^2} + C}

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int \arccos u \mathrm{d}u = u \arccos u - \sqrt {1 - u^2} + C}

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int \arctan u \mathrm{d}u = u \arctan u - \ln \sqrt {1 + u^2} + C}

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int \arccot u \mathrm{d}u = u \arccot u + \ln \sqrt {1 + u^2} + C}

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int \arcsec u \mathrm{d}u = u \arcsec u + \ln \left \vert u + \sqrt {u^2 -1} \right\vert + C}

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int \arccsc u \mathrm{d}u = u \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