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

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

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

prove

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

True

Solution steps

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

Manipulating left side

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

Use the negative angle identity:

cos (- x) = cos (x)

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

Rewrite using trig identities

(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)

= sin^{2} (x)

We showed that the two sides could take the same form

\Rightarrow True

p ro v e 1 + sin (2 θ) = (sin (θ) + cos (θ)) 2

p ro v e tan^{2} (x) - tan (x) + 1 = sec^{2} (x) - tan (x)

p ro v e 2 \cdot cos (x) \cdot sin (x) = sin (2 x)

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

p ro v e \frac{1 + sec ( θ )}{sec ^{2} ( θ )} = 1 + cos^{2} (θ)

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