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

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

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

prove

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

True

Solution steps

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

Manipulating left side

cos (4 x)

Rewrite using trig identities

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

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

We showed that the two sides could take the same form

\Rightarrow True

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

\frac{sec ( x )}{csc ( x )} = tan (x)

\frac{csc ( x )}{sec ( x )} = cot (x)

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

sin (2 A) = 2 sin (A)

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