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To solve the problem (likely the product of sines or cosines as common classic problems), let’s take the example of (\sin10^\circ \sin30^\circ \sin50^\circ \sin70^\circ) (or its equivalent cosine product) and derive the result step-by-step:

Step 1: Simplify using known values and identities

First, note:

• (\sin30^\circ = \frac{1}{2})
• (\sin x = \cos(90^\circ - x)), so:
  (\sin10^\circ = \cos80^\circ), (\sin50^\circ = \cos40^\circ), (\sin70^\circ = \cos20^\circ)

Thus, the product becomes:
[P = \frac{1}{2} \times \cos20^\circ \cos40^\circ \cos80^\circ]

Step 2: Apply the double-angle formula repeatedly

We use (2\sin A \cos A = \sin2A). Multiply numerator and denominator of the cosine product by (2\sin20^\circ):
[ \cos20^\circ \cos40^\circ \cos80^\circ = \frac{2\sin20^\circ \cos20^\circ \cos40^\circ \cos80^\circ}{2\sin20^\circ} ]

Simplify the numerator:
[ \frac{\sin40^\circ \cos40^\circ \cos80^\circ}{2\sin20^\circ} ]

Multiply numerator/denominator by (2) again:
[ \frac{2\sin40^\circ \cos40^\circ \cos80^\circ}{4\sin20^\circ} = \frac{\sin80^\circ \cos80^\circ}{4\sin20^\circ} ]

Multiply numerator/denominator by (2) once more:
[ \frac{2\sin80^\circ \cos80^\circ}{8\sin20^\circ} = \frac{\sin160^\circ}{8\sin20^\circ} ]

Since (\sin160^\circ = \sin(180^\circ - 20^\circ) = \sin20^\circ):
[ \frac{\sin20^\circ}{8\sin20^\circ} = \frac{1}{8} ]

Step3: Compute the final result

[ P = \frac{1}{2} \times \frac{1}{8} = \frac{1}{16} ]

Answer: (\boxed{\dfrac{1}{16}})

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