Is 1-8sin^2(θ)cos^2(θ)=cos(4θ) ?

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

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prove 1-8sin^2(θ)cos^2(θ)=cos(4θ)

prove

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

True

Solution steps

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

Manipulating right 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

\frac{cos ( b )}{sin ( b )} \cdot sin (b) = cos (b)

- 50 cos (4 t) + 50 = 100 sin^{2} (2 t)

\frac{7 sec ( θ ) - 7}{1 - cos ( θ )} = 7 sec (θ)

cos (20 π + x) + cos (21 π - x) = 0

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

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