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

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

Manipulating left side

2 sin^{2} (u)

Rewrite using trig identities

2 sin^{2} (u)

Use the Double Angle identity:

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

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

= - (cos (2 u) - 1)

Simplify

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

- (cos (2 u) - 1)

Distribute parentheses

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

Apply minus-plus rules

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

= - cos (2 u) + 1

= - cos (2 u) + 1

= - cos (2 u) + 1

= 1 - cos (2 u)

We showed that the two sides could take the same form

\Rightarrow 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

p ro v e cos (x) cos (x) = cos^{2} (x)

p ro v e 81 - 9 (3 sin (x))^{2} = - 1

p ro v e csc (x) + 1 = \frac{2}{sin ( x )}

p ro v e \frac{1}{cos ( x )} = arccos (x)

p ro v e tan^{2} (x) sec (x) = cot^{2} (x) csc (x)

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