Is 2sin^2(θ)-1=sin^2(θ)-cos^2(θ) ?

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

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prove 2sin^2(θ)-1=sin^2(θ)-cos^2(θ)

prove

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

True

Solution steps

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

Manipulating right side

sin^{2} (θ) - cos^{2} (θ)

Rewrite using trig identities

= 2 sin^{2} (θ) - 1

We showed that the two sides could take the same form

\Rightarrow True

\frac{1 + cot ( x )}{csc ( x )} = sin (x) + cos (x)

(1 + tan^{2} (x)) cos (x) = sec (x)

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

\frac{tan ( β )}{sec ( β )} = sin (β)

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

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