Question Number 12547 by @ANTARES_VY last updated on 25/Apr/17
$$\boldsymbol{\mathrm{This}} \\ $$$$\boldsymbol{\mathrm{y}}=\boldsymbol{\mathrm{sin}}\frac{\boldsymbol{\mathrm{x}}}{\mathrm{2}}\:\:\boldsymbol{\mathrm{find}}\:\:\boldsymbol{\mathrm{the}}\:\:\boldsymbol{\mathrm{range}}\:\:\boldsymbol{\mathrm{of}}\:\:\boldsymbol{\mathrm{the}}\:\: \\ $$$$\boldsymbol{\mathrm{function}}. \\ $$
Answered by mrW1 last updated on 25/Apr/17
$${x},{y}\in\mathbb{R} \\ $$$$−\mathrm{1}\leqslant{y}\leqslant\mathrm{1} \\ $$
Commented by @ANTARES_VY last updated on 25/Apr/17
$$\boldsymbol{\mathrm{what}}\:\boldsymbol{\mathrm{is}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{soluton}} \\ $$
Commented by mrW1 last updated on 25/Apr/17
$${what}\:{do}\:{you}\:{mean}? \\ $$$${you}\:{wanted}\:{to}\:{know}\:{the}\:{range}\:{of}\:{function}, \\ $$$${I}\:{gave}\:{you}\:{the}\:{answer},\:{that}\:{is} \\ $$$$\:−\mathrm{1}\leqslant{y}\leqslant\mathrm{1}. \\ $$
Commented by ajfour last updated on 26/Apr/17
$$\mathrm{sin}\:\theta=\frac{{p}}{{h}} \\ $$$$−{h}\leqslant{p}\leqslant{h} \\ $$$${so}\:−\mathrm{1}\leqslant\frac{{p}}{{h}}\leqslant\mathrm{1} \\ $$$$\Rightarrow\:−\mathrm{1}\leqslant\mathrm{sin}\:\theta\leqslant\mathrm{1}\:. \\ $$