Respuesta :
We are provided that [tex]{\sf \sin (64^{\circ})=p}[/tex] and we have to find the value of [tex]{\sf 8\sin (16^{\circ})\cdot \cos (16^{\circ})\cdot \cos (32^{\circ})}[/tex] . But let's recall a identity which is gonna be very helpful to do this question i.e
- [tex]{\boxed{\sf 2\sin (\theta)\cos (\theta)=\sin (2\theta)}}[/tex]
Now , coming back on the question ;
[tex]{:\implies \quad \sf 8\sin (16^{\circ})\cdot \cos (16^{\circ})\cdot \cos (32^{\circ})}[/tex]
[tex]{:\implies \quad \sf 4\{ 2\sin (16^{\circ})\cos (16^{\circ})\}\cos (32^{\circ})}[/tex]
Using the above identity ;
[tex]{:\implies \quad \sf 4\sin (32^{\circ})\cos (32^{\circ})}[/tex]
[tex]{:\implies \quad \sf 2\{2\sin (32^{\circ})\cos (32^{\circ})\}}[/tex]
Again using the same identity ;
[tex]{:\implies \quad \sf 2\sin (64^{\circ})}[/tex]
Putting the given value of [tex]{\sf \sin (64^{\circ})}[/tex]
[tex]{:\implies \quad \sf 2p}[/tex]
[tex]{:\implies \quad \bf \therefore \quad \underline{\underline{8\sin (16^{\circ})\cdot \cos (16^{\circ})\cdot \cos (32^{\circ})=2p}}}[/tex]
Proof of the above identity :-
We know that ;
[tex]{\quad \boxed{\sf \sin (A\pm B)=\sin (A)\cos (B)\pm \cos (A)\sin (B)}}[/tex]
So ;
[tex]{:\implies \quad \sf \sin (A+ B)=\sin (A)\cos (B)+\cos (A)\sin (B)}[/tex]
Putting [tex]{\sf A=B=\theta}[/tex] ,
[tex]{:\implies \quad \sf \sin (\theta+ \theta)=\sin (\theta)\cos (\theta)+\cos (\theta)\sin (\theta)}[/tex]
[tex]{:\implies \quad \bf \therefore \quad \underline{\underline{\sin (2\theta)=2\sin (\theta)\cos (\theta)}}}[/tex]
Hence , Proved :D
