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

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

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

prove

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

True

Solution steps

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

Manipulating left side

cos^{2} (x) - sin^{2} (x)

Rewrite using trig identities

cos^{2} (x) - sin^{2} (x)

Use the Pythagorean identity:

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

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

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

Simplify

cos^{2} (x) - (1 - cos^{2} (x)) : 2 cos^{2} (x) - 1

cos^{2} (x) - (1 - cos^{2} (x))

- (1 - cos^{2} (x)) : - 1 + cos^{2} (x)

- (1 - cos^{2} (x))

Distribute parentheses

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

Apply minus-plus rules

- (- a) = a, - (a) = - a

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

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

Simplify

cos^{2} (x) - 1 + cos^{2} (x) : 2 cos^{2} (x) - 1

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

Group like terms

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

Add similar elements:

cos^{2} (x) + cos^{2} (x) = 2 cos^{2} (x)

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

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

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

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

We showed that the two sides could take the same form

\Rightarrow True

sin (16 5^{\circ})

1 - 2 sin (θ) = cos (2 θ)

cos (π)

sec^{2} (1)

x, cos^{2} (x) - sin^{2} (x) = sin (x)

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