This is the final step of one old competition problem, and in textbook it simply says "reader can prove this", but I found this a bit nontrivial.
Show that $\sin^220^\circ\sin40^\circ = \sin10^\circ \sin30^\circ \sin60^\circ$.
I tried a lot of things and formula on this, but never succeeded.
Can someone help?
Best Answer
Trigonometric Proof
This follows almost immediately from $$\sin 20^\circ \sin 40^\circ \sin 80^\circ =\frac{\sqrt 3}{8},\tag{1}$$ which is a corollary of $$\sin x \sin (60^\circ-x) \sin (60^\circ+x)=\frac{\sin(3x)}{4}.$$ My hint is to write one of $\sin 20^\circ$ from $\sin^2 20^\circ$ as $2\sin 10^\circ \cos 10^\circ$, and to note that $\cos 10^\circ=\sin 80^\circ$.
Postscript: We also have $$\cos x\cos(60^\circ-x)\cos(60^\circ+x)=\frac{\cos(3x)}{4},$$ making $$\cos 20^\circ\cos 40^\circ \cos 80^\circ= \frac18.$$ Consequently $$\tan x \tan(60^\circ -x )\tan(60^\circ+x)=\tan (3x)$$ and $$\tan 20^\circ \tan 40^\circ \tan 80^\circ =\sqrt3.$$
We also have the following identity. $$\prod_{j=1}^{n}\sin\left(x+\frac{(j-1)\pi}{n}\right)=\frac{\sin (nx)}{2^{n-1}}$$ for all positive integers $n$. From this identity, $$\prod_{j=1}^{n}\cos\left(x+\frac{(j-1)\pi}{n}\right)=(-1)^{\frac{n-1}{2}}\frac{\cos (nx)}{2^{n-1}}$$ and $$\prod_{j=1}^n\tan\left(x+\frac{(j-1)\pi}{n}\right)=(-1)^{\frac{n-1}{2}}\tan(nx)$$ for all odd positive integers $n$. We also have $$\prod_{j=1}^n\tan\left(x+\frac{(j-1)\pi}{n}\right)=(-1)^{\frac{n}{2}}$$ if $n$ is an even positive integer, which implies $$\prod_{j=1}^{n}\cos\left(x+\frac{(j-1)\pi}{n}\right)=(-1)^{\frac{n}{2}}\frac{\sin (nx)}{2^{n-1}}$$ for each even positive integer $n$.
Geometric Proof
There is also a geometric proof without using $(1)$. Let $ABC$ be a triangle with $\angle BAC=90^\circ$ and $\angle ABC=20^\circ$. Let $E$ and $F$ be internal points of $AC$ and $AB$ such that $BE$ bisects $\angle ABC$ and $\angle ACF=30^\circ$. We want to show that $\angle CFE=20^\circ$.
Note that $\angle BCF=40^\circ$. If $D$ is an internal point of $BF$ such that $CD$ bisects $\angle BDC$, then $\triangle DBC$ is isosceles. Therefore, $DM\perp BC$ if $M$ is the midpoint of $BC$.
By the angular bisector theorem on $\triangle ABC$, $\frac{AE}{CE}=\frac{AB}{BC}$. Since $\triangle MBD\sim \triangle ABC$, we have $$\frac{DM}{BM}=\frac{BC}{AB}=\frac{CE}{AE}.$$ Since $BM=\frac12 BC$, we get $$DB=\frac{CE}{AE} \cdot BM=\frac{BC\cdot CE}{2\cdot AE}.$$
Using the angular bisector theorem with $\triangle FCB$, we get $\frac{DF}{DB}=\frac{CF}{BC}$ so that $$DF=\frac{CF}{BC}\cdot DB=\frac{CE}{AE}\left(\frac{CF}{2}\right).$$ Since $\triangle ACF$ is a right triangle with $\angle ACF=30^\circ$, we get $AF=\frac{CF}{2}$ so that $DF=\frac{CE}{AE}\cdot AF$ or $$\frac{DF}{AF}=\frac{CE}{AE}.$$ This means $EF\parallel CD$. Hence $$\angle CFE=\angle FCD=\frac{1}{2}\angle BCF=20^\circ.$$
From this result, we see using the law of sines on $\triangle CFE$ that $$\frac{CE}{\sin 20^\circ}=\frac{CF}{\sin 130^\circ}=\frac{CF}{\sin 50^\circ}=\frac{CF}{\cos 40^\circ}.$$ Also the law of sines on $\triangle BCF$ yields $$\frac{CF}{\sin 20^\circ}=\frac{BC}{\sin 120^\circ}=\frac{BC}{\sin 60^\circ}.$$ Therefore $$\frac{CE}{\sin 20^\circ}=\frac{BC\sin 20^\circ}{\sin 60^\circ \cos 40^\circ}.\tag{2}$$ But using the law of sines with $\triangle BCE$ gives us $$\frac{CE}{\sin 10^\circ}=\frac{BC}{\sin 100^\circ}=\frac{BC}{\sin 80^\circ}=\frac{BC}{2\sin 40^\circ \cos40^\circ}.\tag{3}$$ From $(2)$ and $(3)$ we obtain $$\frac{BC \sin 10^\circ}{2\sin 40^\circ \cos 40^\circ}= CE=\frac{BC\sin^2 20^\circ}{\sin 60^\circ \cos 40^\circ}.$$ This proves that $$\sin^2 20^\circ \sin 40^\circ =\frac{1}{2}\sin 10^\circ \sin 60^\circ =\sin 10^\circ \sin 30^\circ \sin 60^\circ.$$ And from this result, we can prove $(1)$ as well.