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

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

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

prove

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

True

Solution steps

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

Manipulating right side

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

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

Rewrite using trig identities

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

Use the Double Angle identity:

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

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

Simplify

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

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

Apply the fraction rule:

\frac{- a}{b} = - \frac{a}{b}

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

Remove parentheses:

(a) = a

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

Expand

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

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

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

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

Distribute parentheses

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

Apply minus-plus rules

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

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

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

1 - 1 = 0

= 2 sin^{2} (x)

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

Divide the numbers:

\frac{2}{2} = 1

= - sin^{2} (x)

= - sin^{2} (x)

= - sin^{2} (x)

We showed that the two sides could take the same form

\Rightarrow True

cos (1 \cdot π) = - 1

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

cos (6 0^{\circ} + θ) - cos (6 0^{\circ} - θ) = - 3 sin (θ)

cos (θ) cot (θ) = \frac{1}{sin ( θ )} - sin (θ)

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

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