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

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

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

prove

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

True

Solution steps

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

Manipulating right side

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

Simplify

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

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

Multiply

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

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

Multiply fractions:

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

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

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

Apply rule

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

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

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

Rewrite using trig identities

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

Use the Double Angle identity:

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

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

Simplify

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

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

Expand

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

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

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

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

Distribute parentheses

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

Apply minus-plus rules

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

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

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

1 - 1 = 0

= 2 sin^{2} (t)

= \frac{2 sin ^{2} ( t )}{2}

Divide the numbers:

\frac{2}{2} = 1

= sin^{2} (t)

= sin^{2} (t)

= sin^{2} (t)

We showed that the two sides could take the same form

\Rightarrow True

cos (3 x) = 4 (cos^{3} (x)) - 3 cos (x)

\frac{1 + csc ( x )}{sec ( x )} = cot (x) cos (x)

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

cot (θ) csc (θ) sec (θ) = csc^{2} (θ)

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

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