Is 2sin^2(θ)-1=sin^4(θ)-cos^4(θ) ?

The answer to whether 2sin^2(θ)-1=sin^4(θ)-cos^4(θ) is True

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prove 2sin^2(θ)-1=sin^4(θ)-cos^4(θ)

prove

2 sin^{2} (θ) - 1 = sin^{4} (θ) - cos^{4} (θ)

True

Solution steps

2 sin^{2} (θ) - 1 = sin^{4} (θ) - cos^{4} (θ)

Manipulating right side

sin^{4} (θ) - cos^{4} (θ)

Factor

- cos^{4} (θ) + sin^{4} (θ) : (sin^{2} (θ) + cos^{2} (θ)) (sin (θ) + cos (θ)) (sin (θ) - cos (θ))

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

Rewrite using trig identities

= - 1 + 2 sin^{2} (θ)

= 2 sin^{2} (θ) - 1

We showed that the two sides could take the same form

\Rightarrow True

csc (2 θ) = \frac{1 + cot ^{2} ( θ )}{2 cot ( θ )}

csc^{4} (θ) - cot^{4} (θ) = 2 cot^{2} (θ) + 1

sec (b) cot (b) = csc (b)

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

\frac{cos ( 3 x )}{sin ( x )} + \frac{sin ( 3 x )}{cos ( x )} = 2 cot (2 x)

Is 2sin^2(θ)-1=sin^4(θ)-cos^4(θ) ?
The answer to whether 2sin^2(θ)-1=sin^4(θ)-cos^4(θ) is True