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

cos (x + y) + cos (x - y) = 2 cos (x) cos (y)

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

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

sin (2 x) = \frac{2 tan ( x )}{1 + tan ^{2} ( x )}

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

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