$\large P= sin(20) \times sin(40) \times sin(60) \times sin(80)$

If $P$ can be represented in the form $\large \frac{a}{b}$ where $a$ and $b$ are coprime positive integers

Find $a-b$

Note:All the Angles are in Degree measure.

The answer is -13.

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We can rewrite the expression as

$sin(60)[sin(20) \times sin(40) \times sin(80) = sin(60)[ sin(20) \times sin(60-20) \times sin(60+20)]$ .

Now $sin(\theta) \times sin(60-\theta) \times sin(60+\theta)$ = $\dfrac{sin(3\theta)}{4}$

Then the above statement simplifies to

$sin(60) \times \dfrac{sin(60)}{4}$ which then simplifies to $\boxed{\dfrac{3}{16}}$