Is cos(4B)=1-8sin^2(B)+8sin^4(B) ?

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

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prove cos(4B)=1-8sin^2(B)+8sin^4(B)

prove

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

True

Solution steps

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

Manipulating left side

cos (4 B)

Rewrite using trig identities

= 8 sin^{4} (B) - 8 sin^{2} (B) + 1

= 1 - 8 sin^{2} (B) + 8 sin^{4} (B)

We showed that the two sides could take the same form

\Rightarrow True

tan (θ) sin (θ) cos (θ) = sin^{2} (θ)

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

sin (\frac{π}{2} - u) = cos (u)

csc (θ) = \frac{cot ( θ )}{cos ( θ )}

arccos (x) = \frac{1}{cos ( x )}

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