Prove 8sin^3(x)cos(x)=2sin(2x)-sin(4x)

Answer 1

Prove: 8 sin³(x) cos(x) = 2 sin(2x) - sin(4x)

Prove by starting at:

Left Hand Side (8 sin³(x) cos(x)) and arrive to Right Hand Side (2 sin(2x) - sin(4x))

Known Trigonometry Facts: 2 sin(x) cos(x) = sin(2x)         and      2 sin²(x) = 1 – cos 2x

Left Hand Side:

8 sin³(x) cos(x) = 2*2*2 sin(x) sin(x) sin(x) cos(x)

= 2 * (2 sin(x) cos(x)) * (2 sin²(x))

= 2 * (sin(2x)) * (2 sin²(x))

= 2 sin(2x) * (1 – cos 2x)

= 2 sin(2x) – 2 sin(2x) cos (2x)

= 2 sin(2x) – sin(2(2x))

= 2 sin(2x) – sin(4x)) <== Right Hand Side

Answer 2

8 sin³(x) cos(x) = 2 sin(2x) - sin(4x)

To prove this, you need the double-angle formula:

sin (2θ) = 2 cos (θ) sin (θ)

cos (2θ) = cos²(θ) - sin²(θ)

It's better to manipulate the right side of the equation and let's see if it is equal on the left side:

2 sin(2x) - sin(4x)

Using the double angle formula for sin (2x) and sin (4x):

= 2 * 2 sin (x) cos (x) - 2 sin (2x) cos (2x)

Using double angle formula again for sin (2x) and cos (2x)

= 4 sin (x) cos (x) - 2 * 2 sin (x) cos (x) (cos²(x) - sin²(x))

= 4 sin (x) cos (x) - 4 sin (x) cos (x) (cos²(x) - sin²(x))

Distribute - 4 sin (x) cos (x):

= 4 sin (x) cos (x) - 4 sin(x)cos³(x) + 4sin³(x)cos(x)

Factor out 4 sin (x) cos (x):

= 4 sin (x) cos (x) (1 - cos²(x) + sin²(x))

Using Pythagorean Identity, 1-cos²(x) =sin²(x):

= 4 sin (x) cos (x) (sin²(x) + sin²(x))

Combine like terms:

= 4 sin (x) cos (x) (2 sin²(x))

Perform multiplication:

= 8 sin³(x) cos (x)