Is sin^2(θ)-cos^2(θ)=2sin^2(θ)-1 ?

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

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prove sin^2(θ)-cos^2(θ)=2sin^2(θ)-1

prove

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

True

Solution steps

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

Manipulating left side

sin^{2} (θ) - cos^{2} (θ)

Rewrite using trig identities

sin^{2} (θ) - cos^{2} (θ)

Use the Pythagorean identity:

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

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

= sin^{2} (θ) - (1 - sin^{2} (θ))

Simplify

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

sin^{2} (θ) - (1 - sin^{2} (θ))

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

- (1 - sin^{2} (θ))

Distribute parentheses

= - (1) - (- sin^{2} (θ))

Apply minus-plus rules

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

= - 1 + sin^{2} (θ)

= sin^{2} (θ) - 1 + sin^{2} (θ)

Simplify

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

sin^{2} (θ) - 1 + sin^{2} (θ)

Group like terms

= sin^{2} (θ) + sin^{2} (θ) - 1

Add similar elements:

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

= 2 sin^{2} (θ) - 1

= 2 sin^{2} (θ) - 1

= 2 sin^{2} (θ) - 1

= 2 sin^{2} (θ) - 1

We showed that the two sides could take the same form

\Rightarrow True

cot (\frac{π}{2} - u) = tan (u)

\frac{sec ^{4} ( x ) - 1}{tan ^{2} ( x )} = tan^{2} (x) + 2

\frac{1}{sec ^{2} ( x )} = cos^{2} (x)

tan (x) + cot (x) = \frac{1}{sin ( x ) cos ( x )}

\frac{cot ^{2} ( x )}{csc ( x ) - 1} = csc (x) + 1

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