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

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

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

prove

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

True

Solution steps

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

Manipulating right side

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

Rewrite using trig identities

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

Use the Double Angle identity:

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

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

Simplify

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

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

Multiply fractions:

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

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

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

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

Multiply:

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

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

Remove parentheses:

(a) = a

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

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

cot (- \frac{2 π}{3})

cos (18 0^{\circ})

tan (\frac{π}{2})

sin (- \frac{π}{4})

tan (\frac{π}{3})

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