Is tan(θ/2)=(1-cos(θ))/(sin(θ)) ?

The answer to whether tan(θ/2)=(1-cos(θ))/(sin(θ)) is True

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prove tan(θ/2)=(1-cos(θ))/(sin(θ))

prove

tan (\frac{θ}{2}) = \frac{1 - cos ( θ )}{sin ( θ )}

True

Solution steps

tan (\frac{θ}{2}) = \frac{1 - cos ( θ )}{sin ( θ )}

Let:

u = \frac{θ}{2}

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

Prove

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

True

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

Manipulating right side

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

Rewrite using trig identities

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

Use the Double Angle identity:

sin (2 x) = 2 sin (x) cos (x)

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

Use the Double Angle identity:

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

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

Simplify

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

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

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 sin ( u ) cos ( u )}

Divide the numbers:

\frac{2}{2} = 1

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

Cancel the common factor:

sin (u)

= \frac{sin ( u )}{cos ( u )}

= \frac{sin ( u )}{cos ( u )}

= \frac{sin ( u )}{cos ( u )}

Use the basic trigonometric identity:

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

= tan (u)

We showed that the two sides could take the same form

\Rightarrow True

Therefore

tan (\frac{θ}{2}) = \frac{1 - cos ( θ )}{sin ( θ )}

We showed that the two sides could take the same form

\Rightarrow True

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