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

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