Explanation:

To prove the equation 4sinθsin2θsin4θ = sin3θ, we will use trigonometric identities and simplification techniques. Let's break down the process into smaller steps.

Step 1: Rewrite the equation using trigonometric identities.

The trigonometric identity to be used is:

sin2θ = 2sinθcosθ

By substituting this identity in the equation, we get:

4sinθ(2sinθcosθ)(sin4θ) = sin3θ

Step 2: Simplify the equation.

To simplify the equation further, we can use another trigonometric identity:

sin4θ = 2sin2θcos2θ

Substituting this identity into the equation, we have:

4sinθ(2sinθcosθ)(2sin2θcos2θ) = sin3θ

Step 3: Expand and simplify the equation.

Expanding the equation, we get:

4(2sinθcosθ)(2sinθcosθ)(2sin2θcos2θ) = sin3θ

Simplifying the expression, we have:

16sinθcosθsinθcosθsin2θcos2θ = sin3θ

Step 4: Simplify further using trigonometric identities.

We can use the trigonometric identity:

sin2θ = 2sinθcosθ

By substituting this identity in the equation, we get:

16sinθcosθsinθcosθ(2sinθcosθ)(2(2sinθcosθ)cos2θ) = sin3θ

Simplifying the expression, we have:

16sinθcosθsinθcosθ(2sinθcosθ)(4sinθcos2θ) = sin3θ

Step 5: Simplify and combine like terms.

Multiplying the terms together, we get:

128sin5θcos3θ = sin3θ

Step 6: Cancel out common factors.

Since sin5θ and sin3θ are equal, we can cancel them out:

128cos3θ = 1

Step 7: Solve for θ.

To solve for θ, we divide both sides of the equation by 128:

cos3θ = 1/128

Taking the cube root of both sides, we get:

θ = arccos(1/128)

Thus, we have proved that the equation 4sinθsin2θsin4θ = sin3θ holds true, and the value of θ is given by θ = arccos(1/128).