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

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)

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

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

tan (2 θ) = \frac{2 tan ( θ )}{( 1 - tan ( 2 θ ) )}

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