$\textbf{If $\mathbf{\sqrt{9-8.sin\;50}=a+b.cosec\;50}$, Where $\mathbf{a,b\in\mathbb{Z}}$.Then find $\mathbf{ab=}$$}
341ab=-3?
We have \sin 50^{\circ} \sqrt{9-8 \sin 50^{\circ}} = \sqrt{9 \sin^2 50^{\circ}-8 \sin^3 50^{\circ}}
But
3\sin 50^{\circ} - 4\sin^3 50^{\circ}= \frac{1}{2}
Hence 8\sin^3 50^{\circ}= 6\sin 50^{\circ} - 1
\therefore \ \sqrt{9 \sin^2 50^{\circ}-8 \sin^3 50^{\circ}} = \sqrt{9 \sin^2 50^{\circ} -6 \sin 50^{\circ}+1} = 3 \sin 50^{\circ}-1
(since 3 \sin 50^{\circ}> 3 \sin 50^{\circ}> 3 \sin 45^{\circ}>1
Hence a=3, b=-1
341I looked up at mathlinks.ro but this problem doesnt appear in 2010
62Great Proof again Prophet sir..
I keep wondering as always [1]
am not so frequent at Mathlink but i know the reply you are going to come up is that i should visit mathlink to see brains :D
1708Yes Nishant Sir.
I am also waiting for hsbhatt Sir (Elegant) Solution.
thanks bhatt Sir for Nice Solution.