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

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

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

prove

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

True

Solution steps

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

Manipulating right side

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

Simplify

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

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

Multiply

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

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

Multiply fractions:

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

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

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

Apply rule

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

= \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}

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

sec (t) (csc (t) (tan (t) + cot (t))) = sec^{2} (t) + csc^{2} (t)

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

2 sin^{2} (x) + sin (x) = 0

cot (6 0^{\circ}) = \frac{cos ( 6 0 ^{\circ} )}{sin ( 6 0 ^{\circ} )}

\frac{sec ( x )}{cos ( x )} = sec^{2} (x)

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