Find `sin(tan^(-1)(x+1)).`

Tuesday, March 15, 2011

Find $\displaystyle{\sin{{\left({{\tan}}^{{-{1}}}{\left({x}+{1}\right)}\right)}}}.$

Hello!

The domain of this function is all real numbers, because $\displaystyle{{\tan}}^{{-{1}}}{\left({x}\right)},$ and therefore $\displaystyle{{\tan}}^{{-{1}}}{\left({x}+{1}\right)},$ is defined for all $\displaystyle{x}\in\mathbb{R}.$

By the definition, $\displaystyle{{\tan}}^{{-{1}}}{\left({y}\right)}$ is the number $\displaystyle{w}\in{\left(-\frac{\pi}{{2}},\frac{\pi}{{2}}\right)}$ such that $\displaystyle{y}={\tan{{\left({w}\right)}}}.$ Knowing $\displaystyle{\tan{{\left({w}\right)}}},$ we can find $\displaystyle{\sin{{\left({w}\right)}}}={\sin{{\left({{\tan}}^{{-{1}}}{\left({y}\right)}\right)}}},$

namely

$\displaystyle\frac{{{{\tan}}^{{2}}{\left({w}\right)}}}{{{1}+{{\tan}}^{{2}}{\left({w}\right)}}}=\frac{{\frac{{{{\sin}}^{{2}}{\left({w}\right)}}}{{{{\cos}}^{{2}}{\left({w}\right)}}}}}{{{1}+\frac{{{{\sin}}^{{2}}{\left({w}\right)}}}{{{{\cos}}^{{2}}{\left({w}\right)}}}}}={{\sin}}^{{2}}{\left({w}\right)}.$

This way $\displaystyle{\sin{{\left({w}\right)}}}=\pm\frac{{{\tan{{\left({w}\right)}}}}}{\sqrt{{{1}+{{\tan}}^{{2}}{\left({w}\right)}}}}{\left(=\pm\frac{{y}}{\sqrt{{{1}+{{y}}^{{2}}}}}\right)}.$

Note that for $\displaystyle{w}\in{\left(-\frac{\pi}{{2}},\frac{\pi}{{2}}\right)}$ $\displaystyle{\cos{{\left({w}\right)}}}\gt{0},$ hence the signs of $\displaystyle{\sin{{\left({w}\right)}}}$ and $\displaystyle{\tan{{\left({w}\right)}}}$ are the same. The formula becomes $\displaystyle{\sin{{\left({w}\right)}}}=\frac{{{\tan{{\left({w}\right)}}}}}{\sqrt{{{1}+{{\tan}}^{{2}}{\left({w}\right)}}}}.$