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

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

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

prove

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

True

Solution steps

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

Let:

u = \frac{x}{2}

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

Prove

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

True

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

Manipulating right side

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Simplify

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

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

Factor

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

1 - cos^{2} (2 u)

Factor out common term

- 1

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

Factor

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

cos^{2} (2 u) - 1

Rewrite

1

1^{2}

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

Apply Difference of Two Squares Formula:

x^{2} - y^{2} = (x + y) (x - y)

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

= (cos (2 u) + 1) (cos (2 u) - 1)

= - (cos (2 u) + 1) (cos (2 u) - 1)

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

Cancel

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

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

- cos (2 u) + 1 = - (cos (2 u) - 1)

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

Refine

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

Cancel the common factor:

cos (2 u) - 1

= cos (2 u) + 1

= cos (2 u) + 1

= cos (2 u) + 1

Use the Double Angle identity:

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

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

Simplify

= 2 cos^{2} (u)

= 2 cos^{2} (u)

We showed that the two sides could take the same form

\Rightarrow True

Therefore

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

We showed that the two sides could take the same form

\Rightarrow True

\frac{cot ( θ ) tan ( θ )}{sec ( θ )} = cos (θ)

sin (a) cos (a) = cos^{2} (a) tan (a)

(cos (x)) (tan (x)) (csc (x)) = 1

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

sin (- \frac{π}{4}) = - sin (\frac{π}{4})

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