Is 1-2sin^2(θ/2)=2cos^2(θ/2)-1 ?

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

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prove 1-2sin^2(θ/2)=2cos^2(θ/2)-1

prove

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

True

Solution steps

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

Let:

u = \frac{θ}{2}

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

Prove

1 - 2 sin^{2} (u) = 2 cos^{2} (u) - 1 :

True

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

Manipulating left side

1 - 2 sin^{2} (u)

Rewrite using trig identities

1 - 2 sin^{2} (u)

Use the Pythagorean identity:

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

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

= 1 - 2 (1 - cos^{2} (u))

Simplify

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

1 - 2 (1 - cos^{2} (u))

Expand

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

- 2 (1 - cos^{2} (u))

Apply the distributive law:

a (b - c) = ab - a c

a = - 2, b = 1, c = cos^{2} (u)

= - 2 \cdot 1 - (- 2) cos^{2} (u)

Apply minus-plus rules

- (- a) = a

= - 2 \cdot 1 + 2 cos^{2} (u)

Multiply the numbers:

2 \cdot 1 = 2

= - 2 + 2 cos^{2} (u)

= 1 - 2 + 2 cos^{2} (u)

Subtract the numbers:

1 - 2 = - 1

= 2 cos^{2} (u) - 1

= 2 cos^{2} (u) - 1

= 2 cos^{2} (u) - 1

We showed that the two sides could take the same form

\Rightarrow True

Therefore

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

We showed that the two sides could take the same form

\Rightarrow True

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

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

\frac{tan ( x )}{sec ( x ) - cos ( x )} = csc (x)

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

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

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