Can anyone help with this Trigonometry problem?

Question

I just need to know the first step of how to do this proof. Please help me!

(1+cos x)(1-cos x) = sin2 x

IMPORTANT sin2 is really sine squared but I couldn't type an exponent

Answer 1

The proof is based on the identity

1 = cos^2x + sin^2x

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

1 - cosx + cosx - cos^2x =

1 - cos^2x =

(cos^2x + sin^2x) - cos^2 = (apply the identity)

sin^2 QED

Answer 2

Resolve this problem with factorization:

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

then:

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

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

Resolved now by identity trigonometry
Sin^2 x + Cos^2 x = 1

Sin^2 x = 1 - Cos^2 x

Then:

sin^2 x = sin^2 x

Answer 3

(1 + cos x) (1 - cos x) = sin^2 x
Use the foil method from algebra
(1 - cos x + cos x - cos^2 x) = sin^2 x
1 - cos^2 x = sin^2 x
This is one of the properties of trig functions

Answer 4

Multiply the expressions on the left using the FOIL method. You will get this:

1 - cos(x) + cos(x) - cos^2(x).

Continue from there.

Answer 5

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