How to prove $\sin(9^\circ) \sin(27^\circ) \sin(63^\circ) \sin(81^\circ)=\frac{1}{16}$

Question

How can one prove the following equation?

$$\sin(9^\circ) \sin(27^\circ) \sin(63^\circ) \sin(81^\circ)=\frac{1}{16}$$

Answer 1

Use the fact that $\sin(90^\circ -a)=\cos (a)$ and $\sin (a)\cos (a)=\tfrac12\sin(2a)$:

$$\sin(9^\circ) \cos(9^\circ)\sin(27^\circ)\cos(27^\circ)$$ $$\tfrac14 \sin(18^\circ)\sin(54^\circ)$$

Use a formula $\sin (a)\sin(b)=\tfrac12(\cos(a-b)-\cos(a+b))$ to get

$$\tfrac18(\cos(36^\circ)-\cos(72^\circ))$$

$$\tfrac18(\cos(36^\circ)-\cos(36^\circ)^2+\sin(36^\circ)^2)$$

Here use the values of sine and cosine of $36^\circ$.

Answer 2

hint...the expression is equivalent to $$\cos 9 \sin9\cos27\sin27=\frac 14 \sin 18\sin54=\frac 18(\cos36-\cos72)$$

Then, to obtain an exact expression for $\cos 72$ consider the equation $$\cos3\theta=\cos2\theta$$ which can be solved directly or as a polynomial which can be factorised. The quadratic factor provides the surd form for $\cos 72$...

Answer 3

Let $\sin5y=\sin45^\circ$

$\implies5y=180^\circ n+(-1)^n45^\circ$ where $n$ is any integer

$y=72^\circ m+9^\circ$ where $-2\le m\le2$

But $\sin5y=16\sin^5y − 20\sin^3y + 5\sin y$

So, the roots of $$16\sin^5y − 20\sin^3y + 5\sin y-\sin45^\circ=0$$

are $y=72^\circ m+9^\circ$ where $-2\le m\le2$

$\implies\prod_{m=-2}^2\sin(72^\circ m+9^\circ)=\dfrac{\sin45^\circ}{16}$

Now use $\sin(-A)=-\sin A,\sin(180^\circ-B)=\sin B$

See also :this