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