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

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

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

prove

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

True

Solution steps

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

Manipulating right side

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

Rewrite using trig identities

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

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

Use the Double Angle identity:

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

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

Simplify

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

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

Multiply fractions:

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

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

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

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

Multiply:

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

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

Remove parentheses:

(a) = a

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

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

Divide the numbers:

\frac{2}{2} = 1

= 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

\frac{sin ( θ ) tan ( θ )}{1 - cos ( θ )} = 1 + sec (θ)

sec (t) - \frac{cos ( t )}{1 + sin ( t )} = tan (t)

sec (x) - cos (x) = sin (x) \cdot tan (x)

sin (2 θ) - tan (θ) = tan (θ) cos (2 θ)

tan (\frac{x}{2}) = \frac{tan ( x )}{sec ( x ) + 1}

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