Is 1-2sin^2(θ)=2cos^2(θ)-1 ?

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

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prove 1-2sin^2(θ)=2cos^2(θ)-1

prove

1 - 2 sin^{2} (θ) = 2 cos^{2} (θ) - 1

True

Solution steps

1 - 2 sin^{2} (θ) = 2 cos^{2} (θ) - 1

Manipulating left side

1 - 2 sin^{2} (θ)

Rewrite using trig identities

1 - 2 sin^{2} (θ)

Use the Pythagorean identity:

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

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

= 1 - 2 (1 - cos^{2} (θ))

Simplify

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

1 - 2 (1 - cos^{2} (θ))

Expand

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

- 2 (1 - cos^{2} (θ))

Apply the distributive law:

a (b - c) = ab - a c

a = - 2, b = 1, c = cos^{2} (θ)

= - 2 \cdot 1 - (- 2) cos^{2} (θ)

Apply minus-plus rules

- (- a) = a

= - 2 \cdot 1 + 2 cos^{2} (θ)

Multiply the numbers:

2 \cdot 1 = 2

= - 2 + 2 cos^{2} (θ)

= 1 - 2 + 2 cos^{2} (θ)

Subtract the numbers:

1 - 2 = - 1

= 2 cos^{2} (θ) - 1

= 2 cos^{2} (θ) - 1

= 2 cos^{2} (θ) - 1

We showed that the two sides could take the same form

\Rightarrow True

sin (\frac{π}{4}) = cos (\frac{π}{4})

sec^{2} (x) cot^{2} (x) - 1 = cot^{2} (x)

sin (3 a) + sin (a) = 4 sin (a) - 4 sin (3 a)

tan (β) = \frac{cos ( β )}{sin ( β ) cot ^{2} ( β )}

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

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