Question Number 7200 by Tawakalitu. last updated on 16/Aug/16
$${Prove}\:{that}\: \\ $$$$\frac{{sin}\left(\frac{\Theta}{\mathrm{2}}\right)}{\left(\frac{\Theta}{\mathrm{2}}\right)}\:\:=\:\:\mathrm{1} \\ $$
Commented by Rasheed Soomro last updated on 16/Aug/16
$${Do}\:{you}\:{mean}\:\underset{\Theta\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{sin}\left(\frac{\Theta}{\mathrm{2}}\right)\:}{\frac{\Theta}{\mathrm{2}}}=\mathrm{1}\:? \\ $$
Commented by Tawakalitu. last updated on 16/Aug/16
$${yes}\:{sir} \\ $$
Commented by FilupSmith last updated on 17/Aug/16
$${L}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{sin}\:{x}}{{x}} \\ $$$${L}'{Hopital}'{s}\:{Law} \\ $$$${L}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{cos}\:{x}}{\mathrm{1}} \\ $$$${L}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\mathrm{cos}\left(\mathrm{0}\right) \\ $$$${L}=\mathrm{1} \\ $$
Commented by Tawakalitu. last updated on 17/Aug/16
$${Thank}\:{you}\:{for}\:{your}\:{help}. \\ $$
Answered by Rasheed Soomro last updated on 16/Aug/16
$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\underset{\Theta\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{sin}\:\left(\frac{\Theta}{\mathrm{2}}\right)}{\frac{\Theta}{\mathrm{2}}}=\mathrm{1}\: \\ $$$$−−−−−−−−−−−−−−−− \\ $$$$\:\:\:\:\:\:{Let}\:\frac{\Theta}{\mathrm{2}}=\alpha,\:{then}\:\:{as}\:\Theta\rightarrow\mathrm{0}\Rightarrow\alpha\rightarrow\mathrm{0} \\ $$$$\:\:\:\:\:\underset{\Theta\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{sin}\:\left(\frac{\Theta}{\mathrm{2}}\right)}{\frac{\Theta}{\mathrm{2}}}=\mathrm{1}\Rightarrow\:\:\:\:\underset{\alpha\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{sin}\:\alpha}{\alpha}=\mathrm{1} \\ $$$${Which}\:{is}\:{a}\:{well}\:{known}\:{theorm}\:{of}\:{calculus}. \\ $$$${Hence}\:{proved}. \\ $$
Commented by Tawakalitu. last updated on 16/Aug/16
$${Thank}\:{you}\:{very}\:{much}.\:{i}\:{appreciate}. \\ $$