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

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

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

prove

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

True

Solution steps

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

Manipulating left side

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Simplify

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

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

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

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

Distribute parentheses

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

Apply minus-plus rules

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

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

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

Simplify

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

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

Group like terms

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

Add similar elements:

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

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

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

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

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

We showed that the two sides could take the same form

\Rightarrow True

csc^{2} (t) tan^{2} (t) - 1 = tan^{2} (t)

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

\frac{sec ^{2} ( θ ) - 1}{sin ^{2} ( θ )} = sec^{2} (θ)

sec (x) tan (x) = - tan (x) sec (x)

\frac{1}{csc ( Y ) - cot ( Y )} = csc (Y) + cot (Y)

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