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

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

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

prove

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

True

Solution steps

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

Manipulating right side

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

Rewrite using trig identities

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

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

Use the Double Angle identity:

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

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

Simplify

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

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

Expand

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

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

Apply the distributive law:

a (b - c) = ab - a c

a = - \frac{1}{2}, b = 1, c = 2 sin^{2} (2 x)

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

Apply minus-plus rules

- (- a) = a

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

Simplify

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

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

1 \cdot \frac{1}{2} = \frac{1}{2}

1 \cdot \frac{1}{2}

Multiply:

1 \cdot \frac{1}{2} = \frac{1}{2}

= \frac{1}{2}

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

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

Multiply fractions:

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

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

Cancel the common factor:

2

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

Multiply:

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

= sin^{2} (2 x)

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

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

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

Add similar elements:

\frac{1}{2} - \frac{1}{2} = 0

= sin^{2} (2 x)

= sin^{2} (2 x)

= sin^{2} (2 x)

We showed that the two sides could take the same form

\Rightarrow True

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