Is 2sin^2((9θ)/2)=1-cos(9θ) ?

The answer to whether 2sin^2((9θ)/2)=1-cos(9θ) is True

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prove 2sin^2((9θ)/2)=1-cos(9θ)

prove

2 sin^{2} (\frac{9 θ}{2}) = 1 - cos (9 θ)

True

Solution steps

2 sin^{2} (\frac{9 θ}{2}) = 1 - cos (9 θ)

Let:

u = \frac{9 θ}{2}

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

Prove

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

True

Therefore

2 sin^{2} (\frac{9 θ}{2}) = 1 - cos (9 θ)

We showed that the two sides could take the same form

\Rightarrow True

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

81 - 9 (3 sin (x))^{2} = - 1

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

csc (x) + 1 = \frac{2}{sin ( x )}

csc (x) = - \frac{5}{4}

Is 2sin^2((9θ)/2)=1-cos(9θ) ?
The answer to whether 2sin^2((9θ)/2)=1-cos(9θ) is True