# I wish to prove cos^2x-sin^2x=2cos^2x-1.

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We have to prove that: (cos x)^2 - (sin x)^2 = 2*(cos x)^2 - 1

Starting with the left hand side:

We use the relation (cos x)^2 + (sin x)^2 = 1 or (sin x)^2 = 1 - (cos x)^2

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

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

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

=> 2(cos x)^2 - 1

which is the right hand side.

**Therefore we prove that (cos x)^2 - (sin x)^2 = 2*(cos x)^2 - 1**

You can manipulate only the left side of the equation.

For this reason, you'll use the Pythagorean identity:

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

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

You'll substitute (sin x)^2 to the left side of the equation:

(cos x)^2 - [1 - (cos x)^2] = 2 (cos x)^2 - 1

We'll remove the brackets to the left side:

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

We'll combine like terms and we notice that LHS = RHS:

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