Is cos^2(θ)-sin^2(θ)=2cos^2(θ)-1 ?

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

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prove cos^2(θ)-sin^2(θ)=2cos^2(θ)-1

prove

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

True

Solution steps

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

Manipulating left side

cos^{2} (θ) - sin^{2} (θ)

Rewrite using trig identities

cos^{2} (θ) - sin^{2} (θ)

Use the Pythagorean identity:

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

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

= cos^{2} (θ) - (1 - cos^{2} (θ))

Simplify

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

cos^{2} (θ) - (1 - cos^{2} (θ))

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

- (1 - cos^{2} (θ))

Distribute parentheses

= - (1) - (- cos^{2} (θ))

Apply minus-plus rules

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

= - 1 + cos^{2} (θ)

= cos^{2} (θ) - 1 + cos^{2} (θ)

Simplify

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

cos^{2} (θ) - 1 + cos^{2} (θ)

Group like terms

= cos^{2} (θ) + cos^{2} (θ) - 1

Add similar elements:

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

= 2 cos^{2} (θ) - 1

= 2 cos^{2} (θ) - 1

= 2 cos^{2} (θ) - 1

= 2 cos^{2} (θ) - 1

We showed that the two sides could take the same form

\Rightarrow True

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

(csc (θ) - 1) (csc (θ) + 1) = cot^{2} (θ)

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

sin (\frac{3 π}{2} - θ) = - cos (θ)

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