Is (1+cos(2θ))(1-cos(2θ))=sin^2(2θ) ?

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

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

prove

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

True

Solution steps

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

(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

sin (x) - cos (x) = 0

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

\frac{tan ^{4} ( x ) - 1}{tan ^{2} ( x ) - 1} = sec^{2} (x)

csc (β) - csc (β) cos (2 β) = 2 \cdot sin (β)

cos (27 0^{\circ} - θ) = - sin (θ)

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