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

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1. Begin with the left-hand side (LHS) of the equation:
   $$
   \sin(x)\cos(x)\cos(2x)\cos(4x)
   $$

2. Recognize that $$\sin(x)\cos(x)$$ can be written as:
   $$
   \sin(x)\cos(x) = \frac{1}{2}\sin(2x)
   $$
   So the expression becomes:
   $$
   \frac{1}{2}\sin(2x)\cos(2x)\cos(4x)
   $$

3. Next, express $$\sin(2x)\cos(2x)$$ using a double-angle identity:
   $$
   \sin(2x)\cos(2x) = \frac{1}{2}\sin(4x)
   $$
   Now the expression is:
   $$
   \frac{1}{2} \cdot \frac{1}{2}\sin(4x)\cos(4x) = \frac{1}{4}\sin(4x)\cos(4x)
   $$

4. Again, apply the double-angle formula to $$\sin(4x)\cos(4x)$$:
   $$
   \sin(4x)\cos(4x) = \frac{1}{2}\sin(8x)
   $$
   Therefore, the expression simplifies to:
   $$
   \frac{1}{4} \cdot \frac{1}{2}\sin(8x) = \frac{1}{8}\sin(8x)
   $$

5. We have shown that the LHS simplifies exactly to the right-hand side (RHS):
   $$
   \frac{1}{8}\sin(8x)
   $$

Thus, the given equation
$$
\sin(x)\cos(x)\cos(2x)\cos(4x) = \frac{1}{8}\sin(8x)
$$
is an identity that holds true for all values of $$ x $$.
The equation $$ \sin(x)\cos(x)\cos(2x)\cos(4x) = \frac{1}{8}\sin(8x) $$ is an identity that holds true for all values of $$ x $$.

\( \sin (x)\cos (x)\cos (2x)\cos (4x)=\frac{1}{8}\sin (8x) \)

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