Is cos(4θ)=1-8sin^2(θ)cos^2(θ) ?

The answer to whether cos(4θ)=1-8sin^2(θ)cos^2(θ) is True

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prove cos(4θ)=1-8sin^2(θ)cos^2(θ)

prove

cos (4 θ) = 1 - 8 sin^{2} (θ) cos^{2} (θ)

True

Solution steps

cos (4 θ) = 1 - 8 sin^{2} (θ) cos^{2} (θ)

Manipulating left side

cos (4 θ)

Rewrite using trig identities

= 1 - 8 sin^{2} (θ) cos^{2} (θ)

We showed that the two sides could take the same form

\Rightarrow True

tan^{2} (x) - sin^{2} (x) = (tan (x) sin (x))^{2}

cos^{2} (θ) = sin^{2} (θ) - 1

\frac{sec ( θ )}{sec ( θ ) - 1} = \frac{1}{1 - cos ( θ )}

2 cos (θ) - \frac{cos ( 2 θ )}{cos ( θ )} = sec (θ)

cos (2 \cdot 30) = cos^{2} (30) - sin^{2} (30)

Is cos(4θ)=1-8sin^2(θ)cos^2(θ) ?
The answer to whether cos(4θ)=1-8sin^2(θ)cos^2(θ) is True