Question

If sin 64° = p, determine the value of the following in terms of p.

8 sin 16º.cos 16º.cos32°

Answer 1

We are provided that
`{\sf \sin (64^(\circ))=p}` and we have to find the value of
`{\sf 8\sin (16^(\circ))\cdot \cos (16^(\circ))\cdot \cos (32^(\circ))}` . But let's recall a identity which is gonna be very helpful to do this question i.e

• 
  `{\boxed{\sf 2\sin (\theta)\cos (\theta)=\sin (2\theta)}}`

Now , coming back on the question ;

`{:\implies \quad \sf 8\sin (16^(\circ))\cdot \cos (16^(\circ))\cdot \cos (32^(\circ))}`

`{:\implies \quad \sf 4\{ 2\sin (16^(\circ))\cos (16^(\circ))\}\cos (32^(\circ))}`

Using the above identity ;

`{:\implies \quad \sf 4\sin (32^(\circ))\cos (32^(\circ))}`

`{:\implies \quad \sf 2\{2\sin (32^(\circ))\cos (32^(\circ))\}}`

Again using the same identity ;

`{:\implies \quad \sf 2\sin (64^(\circ))}`

Putting the given value of
`{\sf \sin (64^(\circ))}`

`{:\implies \quad \sf 2p}`

`{:\implies \quad \bf \therefore \quad \underline{\underline{8\sin (16^(\circ))\cdot \cos (16^(\circ))\cdot \cos (32^(\circ))=2p}}}`

Proof of the above identity :-

We know that ;

`{\quad \boxed{\sf \sin (A\pm B)=\sin (A)\cos (B)\pm \cos (A)\sin (B)}}`

So ;

`{:\implies \quad \sf \sin (A+ B)=\sin (A)\cos (B)+\cos (A)\sin (B)}`

Putting
`{\sf A=B=\theta}` ,

`{:\implies \quad \sf \sin (\theta+ \theta)=\sin (\theta)\cos (\theta)+\cos (\theta)\sin (\theta)}`

`{:\implies \quad \bf \therefore \quad \underline{\underline{\sin (2\theta)=2\sin (\theta)\cos (\theta)}}}`

Hence , Proved :D