Is sin^3(2x)=(1/2 sin(2x))(1-cos(4x)) ?

The answer to whether sin^3(2x)=(1/2 sin(2x))(1-cos(4x)) is True

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prove sin^3(2x)=(1/2 sin(2x))(1-cos(4x))

prove

sin^{3} (2 x) = (\frac{1}{2} sin (2 x)) (1 - cos (4 x))

True

Solution steps

sin^{3} (2 x) = \frac{1}{2} sin (2 x) (1 - cos (4 x))

Manipulating right side

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

Rewrite using trig identities

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

= \frac{1}{2} sin (2 x) (1 - cos (2 \cdot 2 x))

Use the Double Angle identity:

cos (2 x) = 1 - 2 sin^{2} (x)

= \frac{1}{2} sin (2 x) (1 - (1 - 2 sin^{2} (2 x)))

Simplify

\frac{1}{2} sin (2 x) (1 - (1 - 2 sin^{2} (2 x))) : sin^{3} (2 x)

\frac{1}{2} sin (2 x) (1 - (1 - 2 sin^{2} (2 x)))

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{1 \cdot sin ( 2 x ) ( 1 - ( 1 - 2 sin ^{2} ( 2 x ) ) )}{2}

Multiply:

1 \cdot sin (2 x) = sin (2 x)

= \frac{sin ( 2 x ) ( - ( - 2 sin ^{2} ( 2 x ) + 1 ) + 1 )}{2}

Expand

1 - (1 - 2 sin^{2} (2 x)) : 2 sin^{2} (2 x)

1 - (1 - 2 sin^{2} (2 x))

- (1 - 2 sin^{2} (2 x)) : - 1 + 2 sin^{2} (2 x)

- (1 - 2 sin^{2} (2 x))

Distribute parentheses

= - (1) - (- 2 sin^{2} (2 x))

Apply minus-plus rules

- (- a) = a, - (a) = - a

= - 1 + 2 sin^{2} (2 x)

= 1 - 1 + 2 sin^{2} (2 x)

1 - 1 = 0

= 2 sin^{2} (2 x)

= \frac{2 sin ^{2} ( 2 x ) sin ( 2 x )}{2}

sin (2 x) \cdot 2 sin^{2} (2 x) = 2 sin^{3} (2 x)

sin (2 x) \cdot 2 sin^{2} (2 x)

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

sin (2 x) sin^{2} (2 x) = sin^{1 + 2} (2 x)

= 2 sin^{1 + 2} (2 x)

Add the numbers:

1 + 2 = 3

= 2 sin^{3} (2 x)

= \frac{2 sin ^{3} ( 2 x )}{2}

Divide the numbers:

\frac{2}{2} = 1

= sin^{3} (2 x)

= sin^{3} (2 x)

= sin^{3} (2 x)

We showed that the two sides could take the same form

\Rightarrow True

Is sin^3(2x)=(1/2 sin(2x))(1-cos(4x)) ?
The answer to whether sin^3(2x)=(1/2 sin(2x))(1-cos(4x)) is True