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

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

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

prove

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

True

Solution steps

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

Let:

u = \frac{t}{2}

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

Prove

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

True

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

Manipulating right side

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

Rewrite using trig identities

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

Use the Double Angle identity:

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

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

Simplify

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

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

Expand

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

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

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

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

Distribute parentheses

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

Apply minus-plus rules

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

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

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

1 - 1 = 0

= 2 sin^{2} (u)

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

Divide the numbers:

\frac{2}{2} = 1

= sin^{2} (u)

= sin^{2} (u)

= sin^{2} (u)

We showed that the two sides could take the same form

\Rightarrow True

Therefore

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

We showed that the two sides could take the same form

\Rightarrow True

cos (θ) = (sec (θ) + tan (θ)) (1 - sin (θ))

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

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

\frac{sin ( 6 x ) + sin ( 2 x )}{cos ( 6 x ) - cos ( 2 x )} = - cot (2 x)

(sin (x) + cos (x)) 2 = 1 + sin (2 x)

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