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

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

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

prove

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

True

Solution steps

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

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

(1 + cos (x)) (1 - cos (x))

Expand

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

(1 + cos (x)) (1 - cos (x))

Apply Difference of Two Squares Formula:

(a + b) (a - b) = a^{2} - b^{2}

a = 1, b = cos (x)

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

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - cos^{2} (x)

= 1 - cos^{2} (x)

Use the Pythagorean identity:

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

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

= sin^{2} (x)

\Rightarrow True

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

p ro v e sin (t) csc (t) = 1

p ro v e cos (x) = sin (x + \frac{π}{2})

p ro v e sec (2 θ) = \frac{sec ^{2} ( θ )}{2 - sec ^{2} ( θ )}

p ro v e tan (x) + cot (x) = \frac{2}{sin ( 2 x )}

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