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

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

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

prove

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

True

Solution steps

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

Manipulating right 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

\frac{1 + tan ( θ )}{1 + cot ( θ )} = tan (θ)

cos (x + 1) = cos (- x - 1)

cos (x) (tan (x) sec (x)) = sin (x) - 1

\frac{tan ( θ )}{sin ( θ )} = sec (θ)

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

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