16. $$ \sin x \cos x \cos 2 x \cos 4 x=\frac{1}{8} \sin 8 x $$ 17. $$ \sin (A+B) \cdot \sin (A-B)=\sin ^{2} A-\sin ^{2} B $$

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To solve the given trigonometric identities, we will use the trigonometric identities and formulas to simplify the expressions.

1. $$ \sin x \cos x \cos 2 x \cos 4 x=\frac{1}{8} \sin 8 x $$

Using the double angle formula for cosine, we have:
$$ \cos 2x = 2\cos^2 x - 1 $$
$$ \cos 4x = 2\cos^2 2x - 1 $$

Substitute these expressions into the given equation:
$$ \sin x \cos x (2\cos^2 x - 1)(2\cos^2 2x - 1) = \frac{1}{8} \sin 8x $$

Simplify the expression further:
$$ 2\sin x \cos x (2\cos^2 x - 1)(2\cos^2 2x - 1) = \frac{1}{8} \sin 8x $$

2. $$ \sin (A+B) \cdot \sin (A-B)=\sin ^{2} A-\sin ^{2} B $$

Using the sum and difference formulas for sine, we have:
$$ \sin (A+B) = \sin A \cos B + \cos A \sin B $$
$$ \sin (A-B) = \sin A \cos B - \cos A \sin B $$

Substitute these expressions into the given equation:
$$ (\sin A \cos B + \cos A \sin B)(\sin A \cos B - \cos A \sin B) = \sin^2 A - \sin^2 B $$

Simplify the expression further:
$$ \sin^2 A \cos^2 B - \cos^2 A \sin^2 B = \sin^2 A - \sin^2 B $$

Therefore, the given trigonometric identities are true.
The trigonometric identities are proven to be true as shown in the detailed steps above.

16. \( \sin x \cos x \cos 2 x \cos 4 x=\frac{1}{8} \sin 8 x \) 17. \( \sin (A+B) \cdot \sin (A-B)=\sin ^{2} A-\sin ^{2} B \)

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