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

Another way of solving the problem (cos x)² - (sin x)² = 2(cos x)² - 1

is as follows:

Consider the right-hand side (RHS) of the equation: 2(cos x)² - 1

From the Pythagorean Identity, we know that (cos x)² + (sin x)² = 1

Thus substitute 1 with (cos x)² + (sin x)² in the RHS of the equation:

2 (cos x)² - ((sin x)² + (cos x)² ).

Expand to give 2(cos x)² - (sin x)² - (cos x)² = (cos x)² - (sin x)² (as given by the LHS of the original equation).

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

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Alternatively, we can start with the right-hand side and show that it equals the left hand side.

Rewrite the right hand side, 2(cosx)² - 1, as (cosx)² + (cosx)² - 1, which can also be rewritten as

(cosx)² - (1 - (cosx)²) by switching the order of the second and third terms and factoring out negative sign.

Then, according to Pythagorean Identity, (sinx)² + (cosx)² = 1, the expression in the parenthesis, 1 - (cosx)², can be replaced by (sinx)²:

(cosx)² - (1 - (cosx)²) = (cosx)² - (sinx)².

This, we have shown that (cosx - (sinx = 2(cosx)² - 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