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

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

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

prove

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

True

Solution steps

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

Manipulating left side

2 cos^{2} (w) - 1

Rewrite using trig identities

2 cos^{2} (w) - 1

Use the Pythagorean identity:

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

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

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

Simplify

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

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

Expand

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

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

Apply the distributive law:

a (b - c) = ab - a c

a = 2, b = 1, c = sin^{2} (w)

= 2 \cdot 1 - 2 sin^{2} (w)

Multiply the numbers:

2 \cdot 1 = 2

= 2 - 2 sin^{2} (w)

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

Simplify

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

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

Group like terms

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

Add/Subtract the numbers:

2 - 1 = 1

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

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

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

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

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

We showed that the two sides could take the same form

\Rightarrow True

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

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

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