Prove-that-sin-2-2-1-

Question Number 7200 by Tawakalitu. last updated on 16/Aug/16

P r o v e t h a t

\frac{s i n (\frac{Θ}{2})}{(\frac{Θ}{2})} = 1

Commented by Rasheed Soomro last updated on 16/Aug/16

D o y o u m e a n \lim_{Θ \to 0} \frac{\sin (\frac{Θ}{2})}{\frac{Θ}{2}} = 1 ?

Commented by Tawakalitu. last updated on 16/Aug/16

y e s s i r

Commented by FilupSmith last updated on 17/Aug/16

L = \lim_{x \to 0} \frac{\sin x}{x}

L^{'} {H o p i t a l}^{'} s L a w

L = \lim_{x \to 0} \frac{\cos x}{1}

L = \lim_{x \to 0} \cos (0)

L = 1

Commented by Tawakalitu. last updated on 17/Aug/16

T h a n k y o u f o r y o u r h e l p .

Answered by Rasheed Soomro last updated on 16/Aug/16

\lim_{Θ \to 0} \frac{\sin (\frac{Θ}{2})}{\frac{Θ}{2}} = 1

- - - - - - - - - - - - - - - -

L e t \frac{Θ}{2} = α, t h e n a s Θ \to 0 \Rightarrow α \to 0

\lim_{Θ \to 0} \frac{\sin (\frac{Θ}{2})}{\frac{Θ}{2}} = 1 \Rightarrow \lim_{α \to 0} \frac{\sin α}{α} = 1

W h i c h i s a w e l l k n o w n t h e o r m o f c a l c u l u s .

H e n c e p r o v e d .

Commented by Tawakalitu. last updated on 16/Aug/16

T h a n k y o u v e r y m u c h . i a p p r e c i a t e .