We are given the expression: $\sin 10 ° \sin 50 ° \sin 70 °$ and we need to find its product value.

A useful identity for products of sines is: $\sin A \sin B = \frac{1}{2} [\cos (A - B) - \cos (A + B)]$ . We will apply this step by step.

Let A = 50° and B = 70°.

$\sin 50 ° \sin 70 ° = \frac{1}{2} [\cos (50 ° - 70 °) - \cos (50 ° + 70 °)] = \frac{1}{2} [\cos (- 20 °) - \cos (120 °)]$

Since cosine is an even function, cos(-θ) = cosθ.

$\sin 50 ° \sin 70 ° = \frac{1}{2} [\cos 20 ° - \cos 120 °]$

We know that cos120° = cos(180° - 60°) = -cos60° = -1/2.

$\sin 50 ° \sin 70 ° = \frac{1}{2} [\cos 20 ° - (- \frac{1}{2})] = \frac{1}{2} [\cos 20 ° + \frac{1}{2}] = \frac{1}{2} \cos 20 ° + \frac{1}{4}$

Our original expression now becomes:

$\sin 10 ° \times (\frac{1}{2} \cos 20 ° + \frac{1}{4}) = \frac{1}{2} \sin 10 ° \cos 20 ° + \frac{1}{4} \sin 10 °$

We use the identity: $\sin A \cos B = \frac{1}{2} [\sin (A + B) + \sin (A - B)]$ .

Let A = 10° and B = 20°.

$\sin 10 ° \cos 20 ° = \frac{1}{2} [\sin (10 ° + 20 °) + \sin (10 ° - 20 °)] = \frac{1}{2} [\sin 30 ° + \sin (- 10 °)]$

Since sine is an odd function, sin(-θ) = -sinθ.

$\sin 10 ° \cos 20 ° = \frac{1}{2} [\sin 30 ° - \sin 10 °] = \frac{1}{2} [\frac{1}{2} - \sin 10 °] = \frac{1}{4} - \frac{1}{2} \sin 10 °$

Now plug this result back into our expression from Step 3:

$\frac{1}{2} (\frac{1}{4} - \frac{1}{2} \sin 10 °) + \frac{1}{4} \sin 10 ° = \frac{1}{8} - \frac{1}{4} \sin 10 ° + \frac{1}{4} \sin 10 °$

Notice that the terms $- \frac{1}{4} \sin 10 °$ and $+ \frac{1}{4} \sin 10 °$ cancel each other out.

$\frac{1}{8} - \frac{1}{4} \sin 10 ° + <mn$